Could Euclid have proven that multiplication of real numbers distributes over addition?
I hope it is permissible to assume, as Euclid himself did in the proof of V.18, the existence of a fourth proportional to any three given magnitudes. If not, my third proof (underneath the third horizontal line in this answer) avoids making this assumption, but it succeeds only in proving that if the product of $(c + d)/e$ and $a/b$ exists (note the reversal of the order of the two factors), it equals $(f + g)/h$.
My guess, for what it's worth, is that the existence assumption (or something else strong enough to imply it) cannot be eliminated; but to be sure of this, one would have to write out a theory in full detail, argue convincingly that it represents what Euclid really had in mind, and show that it has a model in which there are magnitudes $a, b, c, d, e, f, g, h$, satisfying the stated conditions, for which the product of $a/b$ and $(c + d)/e$ (or vice versa) does not exist.
(See Heath's commentary on Definitions 9 and 10 of Book V for a discussion of the fact that Euclid himself never defines products, i.e. "compound ratios".)
First proof
I begin by proving a "perturbed" version of the required proposition:
Suppose that the product of $a/b$ and $c/e$ is $f/h$, and the product of $d/b$ and $c/e$ is $g/h$. Then the product of $(a + d)/b$ and $c/e$ is $(f + g)/h$.
By the definition of product, there exist $r, s, t$ such that $a/b = r/s$, $c/e = s/t$, and $f/h = r/t$.
By the assumption of the existence of a fourth proportional, there exists $u$ such that $d/b = u/s$.
Since $d/b = u/s$ and $c/e = s/t$, the product of $d/b$ and $c/e$ is $u/t$.
Hence (by V.11, if we're being fussy), $g/h = u/t$.
We have established $f/h = r/t$ and $g/h = u/t$, whence V.24 gives $(f + g)/h = (r + u)/t$.
We also established $a/b = r/s$ and $d/b = u/s$, whence V.24 gives $(a + d)/b = (r + u)/s$. Because $c/e = s/t$, it follows from the definition of product that the product of $(a + d)/b$ and $c/e$ is $(r + u)/t$.
Hence (by V.11 again), the product of $(a + d)/b$ and $c/e$ is $(f + g)/h$, as required. $\square$
To keep things tidy, I use the assumption of the existence of a fourth proportional again, to package up V.23 in the form of a useful lemma which expresses the commutativity of the product operation.
(We will need the assumption of the existence of a fourth proportional, anyway, if we wish to be able to define the product of any two given ratios - rather than relying on there happening to be magnitudes $r, s, t$ in the required proportions, as in the stated definition - so we may as well make heavy use of it.)
If the product of $u/v$ and $w/x$ is $y/z$, then the product of $w/x$ and $u/v$ is also $y/z$.
By the definition of the product, there exist $r, s, t$ such that $u/v = r/s$, $w/x = s/t$, and $y/z = r/t$.
By the assumption of the existence of a fourth proportional, there exists $q$ such that $q/u = w/x$.
By the definition of the product (and also, strictly speaking, by the reflexivity and symmetry of the relation of equality between ratios: $u/v = u/v$, and $w/x = q/u$), the product of $w/x$ and $u/v$ is $q/v$.
But the magnitudes $r, s, t$, and the "others equal to them in multitude", viz. $q, u, v$, now stand to one another in the relation required by V.23, "perturbed ratios ex aequali". That is, $q/u\ (= w/x) = s/t$ (strictly, we're using V.11 again), and $u/v = r/s$, whence V.23 implies $q/v = r/t$, as required. $\square$
Now we can easily derive the proposition which you actually wanted to prove:
Suppose that the product of $a/b$ and $c/e$ is $f/h$, and the product of $a/b$ and $d/e$ is $g/h$. Then the product of $a/b$ and $(c + d)/e$ is $(f + g)/h$.
By the lemma just proved, the product of $c/e$ and $a/b$ is $f/h$, and the product of $d/e$ and $a/b$ is $g/h$. By the "perturbed" proposition which we proved at the outset, the product of $(c + d)/e$ and $a/b$ is $(f + g)/h$. By the lemma, again, the product of $a/b$ and $(c + d)/e$ is also $(f + g)/h$, as required. $\square$
Second proof
Making even more shameless use of the heinous assumption of the existence of fourth proportionals to any three given magnitudes, we can prove your proposition more directly - essentially by using a form of "scalar multiplication" of magnitudes by ratios:
Lemma 2. If $f/h$ is the product of $a/b$ and $c/e$, then there exists $p$ such that $f/p = a/b$ and $p/h = c/e$.
By the assumption of the existence of fourth proportionals, there exists $p$ such that $p/h = c/e$, and there exists $q$ such that $q/p = a/b$.
By the definition of product, there exist $r, s, t$ such that $s/t = c/e$, $r/s = a/b$, and $r/t = f/h$.
Thus (by V.11, twice) $s/t = p/h$, and $r/s = q/p$. By V.22 (ratios ex aequali), $r/t = q/h$. That is (by V.11 again), $f/h = q/h$. By V.9, $f = q$. Therefore $f/p = a/b$, as required for the lemma. $\square$
Apply Lemma 2 to $a, b, c, e, f, h$ as given, obtaining $p$ as described in the conclusion.
Apply Lemma 2 also to $a, b, d, e, g, h$, respectively, obtaining $u$ such that $g/u = a/b$ and $u/h = d/e$.
We have $f/p = a/b = g/u$, therefore $(f + g)/(p + u) = a/b$, by V.12.
Also $p/h = c/e$ and $u/h = d/e$, therefore $(p + u)/h = (c + d)/e$, by V.24.
Therefore $(f + g)/h$ is the product of $a/b$ and $(c + d)/e$, as required for the given proposition. $\square$
Third proof
Here is a longer proof, which avoids the assumption of the existence of a fourth proportional. It only uses principles that Euclid states explicitly or else consistently takes for granted in the Elements (supplemented by a fairly modern understanding of the properties of the system of positive integers, and, of course, a very convenient modern use of algebraic symbolism):
Lemma 3.1 For all magnitudes $x, y$ (of the same kind), and all positive integers $m, n$, $$ mx \leqslant ny \iff (\forall\, p, q \in \mathbb{N})\, (pn < qm \implies px < qy). $$
Suppose first that the LHS is false, i.e. $mx > ny$. Then, by the Archimedean axiom (expressed as Definition 4 of Book V), there exists $r \in \mathbb{N}$ such that $$ r(mx - ny) > x. $$ By V.5, $r(mx) - r(ny) > x$. That is: $$ rmx - rny > x. $$
For the sake of brevity, I shall make free use of identities such as $r(mx) = (rm)x$, and I will often simply write both sides without parentheses. (Sometimes there will be products of as many as four integers with a given magnitude, and in such cases, the use of a brief notation will be well-nigh essential.)
We have $rmx > x$, therefore $rm > 1$.
By an unstated assumption to the effect that addition of magnitudes preserves order (see e.g.\ Scott's notes - mentioned in another answer - for an attempt to write out all the assumptions and definitions needed), we infer also $$rmx - x > rny. $$ Hence, by V.6, and the fact that $1x = x$, we get $$ (rm - 1)x > rny. $$ Taking $p = rm - 1$, $q = rn$, we get $pn < qm$, but $px > qy$, so the RHS is false.
The proof of the converse - i.e. that the truth of the LHS implies the truth of the RHS - is simpler, and doesn't require the Archimedean axiom. Suppose $mx \leqslant ny$, and $pn < qm$. Then (by V.3, the commutativity of multiplication of positive integers, and various unstated properties of magnitudes - again, see Scott's notes for one way of going about formalising the theory - it's not feasible to spell everything out in an MSE answer!)\ $$ n(px) = (np)x = (pn)x < (qm)x = q(mx) \leqslant q(ny) = (qn)y = (nq)y = n(qy), $$ whence $px < qy$, as required. $\square$
Corollary 3.2 For any magnitudes $x, y$ (of the same kind), and $u, v$ (of the same kind - as each other, not necessarily the same kind as $x, y$), $$ y/x = v/u \iff (\forall m, n \in \mathbb{N})\, (mx < ny \iff mu < nv). $$
That the LHS implies the RHS is immediate from Euclid's (or rather, Eudoxus's) Definition 5. The point is that we can simplify our arguments by using the fact that the truth of the RHS is not only necessary but also sufficient for the truth of the LHS. Suppose, then, that the RHS is true. By Lemma 3.1, \begin{equation*} \begin{split} mx \leqslant ny & \iff (\forall\, p, q \in \mathbb{N})\, (pn < qm \implies px < qy) \\ & \iff (\forall\, p, q \in \mathbb{N})\, (pn < qm \implies pu < qv) \\ & \iff mu \leqslant nv, \end{split} \end{equation*} and the LHS follows from this, in conjunction with the RHS, and the fact that the set of magnitudes of a given kind is totally ordered. $\square$
Proposition 3.3 For magnitudes $x, y$ of the same kind, there exists a unique positive integer $n$ such that $x < ny \leqslant x + y$.
The Archimedean axiom implies that there are values of the positive integer variable $n$ such that $ny > x$. Let $n$ have the least such value. If $n = 1$, then $ny = y < x + y$, so the condition on $n$ is satisfied. Suppose, on the other hand, that $n > 1$. Then $(n - 1)y \leqslant x$, therefore $ny \leqslant x + y$, so the condition on $n$ is satisfied in this case also.
If two distinct multiples of $y$ are both greater than $x$, then the larger of them must be greater than $x + y$, so only one value of $n$ satisfies the condition. $\square$
Very similarly (we will be needing both versions):
Proposition 3.4 For magnitudes $x, y$ of the same kind, there exists a unique positive integer $n$ such that $x \leqslant ny < x + y$.
The Archimedean axiom implies that there are values of the positive integer variable $n$ such that $ny \geqslant x$. Let $n$ have the least such value. If $n = 1$, then $ny = y < x + y$, so the condition on $n$ is satisfied. Suppose, on the other hand, that $n > 1$. Then $(n - 1)y < x$, therefore $ny < x + y$, so the condition on $n$ is satisfied in this case also.
If two distinct multiples of $y$ are both greater than or equal to $x$, then the larger of them must be greater than or equal to $x + y$, so only one value of $n$ satisfies the condition. $\square$
Corollary 3.5 For magnitudes $x, y$ such that $x > y$, there exists a unique positive integer $m$ such that $my < x \leqslant (m + 1)y$.
Because $x > y$, the value of $n$ in Proposition 3.4 cannot be $1$, so we can put $n = m + 1$, and the result of the proposition becomes $x \leqslant (m + 1)y < x + y$. But $(m + 1)y = my + y$, and the inequality $my + y < x + y$ is equivalent to $my < x$. The uniqueness of $m$ follows from that of $n$. $\square$
(Alternatively $my \leqslant x < (m + 1)y$, but we won't be needing this version.)
Lemma 3.6 For magnitudes $x, y, z$ (of the same kind), $$ x < z \iff (\exists\, r, s \in \mathbb{N})\ rx < sy < rz. $$
If the RHS is true, then $rx < rz$, which implies the LHS.
For the proof of the converse implication, suppose that the LHS is true. The Archimedean axiom then implies that there exists $r \in \mathbb{N}$ such that \begin{align*} & {} r(z - x) > y, \\ \therefore \quad & {} rx + y < rz. \end{align*} By Proposition 3.3, there exists a positive integer $s$ such that $$ rx < sy \leqslant rx + y, $$ which, in conjunction with the condition on $r$, implies the conclusion. $\square$
Corollary 3.7 If the product of $c/e$ and $a/b$ is $f/h$, then, for all $n, m \in \mathbb{N}$, $$ mh < nf \iff (\exists\, r, s \in \mathbb{N})\ (rmb < sa \text{ and } se < rnc). $$ Conversely, if this condition holds, then the product of $c/e$ and $a/b$ is $f/h$, assuming only that the product exists.
Suppose first that the product of $c/e$ and $a/b$ is $f/h$. By the definition of product, there exist magnitudes $x, y, z$ (of the same kind) such that $c/e = z/y$, $a/b = y/x$, and $f/h = z/x$. Hence, by Lemma 3.6 and the definition of equal ratios (we don't need to appeal to Corollary 3.2 yet), \begin{align*} mh < nf & \iff mx < nz \\ & \iff (\exists\, r, s \in \mathbb{N})\ (rmx < sy \text{ and } sy < rnz) \\ & \iff (\exists\, r, s \in \mathbb{N})\ (rmb < sa \text{ and } se < rnc). \end{align*}
Conversely, suppose that the stated condition holds. Suppose also that the product of $c/e$ and $a/b$ exists; let it be equal to $f'/h'$. Then, for all $m, n \in \mathbb{N}$, we have $mh < nf \iff mh' < nf'$. By Corollary 3.2, $f/h = f'/h'$. Therefore, the product of $c/e$ and $a/b$ is $f/h$. $\square$
Lemma 3.8 For magnitudes $x, y, z$ (of the same kind), and positive integers $n, m$, $$ mx < n(y + z) \iff (\exists\, r, j, k \in \mathbb{N})\ (rm = j + k,\ jx < rny, \text{ and } kx < rnz). $$
If the RHS is true, then $$ rmx = (j + k)x = jx + kx < rny + rnz = rn(y + z), $$ which implies the LHS.
For the proof of the converse implication, suppose that the LHS is true.
In the same way as in the previous lemmas, the Archimedean axiom implies that there exists $r^* \in \mathbb{N}$ such that $$ r^*(n(y + z) - mx) > 2x. $$ Therefore, for all $r \geqslant r^*$, we have $r(n(y + z) - mx) > 2x$, whence $$ (rm + 2)x < rn(y + z). $$ The same axiom also implies that there exist $r', r'' \in \mathbb{N}$ such that $r'ny > x$ and $r''nz > x$. Let $r = \max\{r^*, r', r'', 2\}$. Because $r \geqslant r'$ and $r \geqslant r''$, we have $rny > x$ and $rnz > x$. By Corollary 3.5, there exist $j', k' \in \mathbb{N}$ such that \begin{gather*} j'x < rny \leqslant (j' + 1)x, \\ k'x < rnz \leqslant (k' + 1)x. \end{gather*} Hence, $$ (rm + 2)x < rn(y + z) = rny + rnz \leqslant (j' + k' + 2)x, $$ implying $rm < j' + k'$. Because $rm \geqslant r \geqslant 2$, and every integer between $2$ and $j' + k'$ (inclusive) is the sum of an integer between $1$ and $j'$ and an integer between $1$ and $k'$, it follows that $rm = j + k$, where $1 \leqslant j \leqslant j'$ and $1 \leqslant k \leqslant k'$. We have $jx \leqslant j'x < rny$ and $kx \leqslant k'x < rnz$, so the RHS is true. $\square$
Theorem 3.9 Suppose that the product of $c/e$ and $a/b$ is $f/h$, and the product of $d/e$ and $a/b$ is $g/h$. Then the product of $(c + d)/e$ and $a/b$, if it exists, is $(f + g)/h$.
By Corollary 3.7, what we have to show is that, for all $n, m \in \mathbb{N}$, $$ mh < n(f + g) \iff (\exists\, p, q \in \mathbb{N})\ (pmb < qa \text{ and } qe < pn(c + d)). $$
Suppose first that $mh < n(f + g)$. By Lemma 3.8, there exist $r, j, k \in \mathbb{N}$ such that $j + k = rm$, $jh < rnf$, and $kh < rng$. Applying Corollary 3.7 to these two inequalities, we find that there are $s, t \in \mathbb{N}$ such that $sjb < ta$ and $te < srnc$, and $u, v \in \mathbb{N}$ such that $ukb < va$ and $ve < urnd$. Hence, \begin{align*} surmb & = sujb + sukb \\ & < tua + sva \\ & = (tu + sv)a, \\ (tu + sv)e & = tue + sve \\ & < surnc + surnd \\ & = surn(c + d). \end{align*} With $p = sur$, $q = tu + sv$, we have $pmb < qa$ and $qe < pn(c + d)$, as required.
To prove the converse implication, suppose that there exist $p, q \in \mathbb{N}$ such that $pmb < qa$ and $qe < pn(c + d)$. Applying Lemma 3.8 to the second inequality, we find that there exist $r, j, k \in \mathbb{N}$ such that $j + k = rq$, $je < rpnc$, and $ke < rpnd$.
Hence, $(pmj)b < (qj)a$ and $(qj)e < (pqrn)c$; and applying Corollary 3.7, we infer $(pmj)h < (pqrn)f$. Similarly, $(pmk)b < (qk)a$ and $(qk)e < (pqrn)d$; and applying Corollary 3.7 again, we infer $(pmk)h < (pqrn)g$. Now we have: \begin{align*} pqrmh & = pm(j + k)h \\ & = pmjh + pmkh \\ & < pqrnf + pqrng \\ & = pqrn(f + g), \end{align*} whence $mh < n(f + g)$, as required. $\square$
The answer is to your question is "yes" provided the system of magnitudes is sufficiently populated. This (and much more!) is all worked out in an old mimeograph by Dana Scott called "General Theory of Magnitudes". I have a scanned copy that I'd be willing to email you if you message me.
The idea of Scott's proof, which has too many details for me to include here, is the following. First, we need to assume that the system of magnitudes is complete:
A system $S'$ of magnitudes is complete if for every other system $S$ of magnitudes and all $x, z\in S$ and $z' \in S'$ there exists an $x' \in S'$ such that $x/z = x'/z'$.
It is possible, and not difficult, but somewhat space consuming, to show the next result. Call a function $f: S \to S'$ between systems of magnitudes additive if $f(x+y) = f(x) \ +' \ f(y)$.
There exists a one-to-one correspondence between ratios of magnitudes in $S$ and additive functions from $S$ into $S$ which preserves addition and multiplication (and inversion and the unit element, when these are appropriately defined).
The reason for invoking this result is that it makes it very easy to verify all sorts of algebraic laws. For example, for distributivity, let $h_{a}: S \to S$, for $a$ a ratio of magnitudes, be the unique additive function determined by the result just stated. Then your result follows by simply checking
$h_{a} \cdot (h_{b} + h_{c}) = h_{a} \cdot h_{b} + h_{a} \cdot h_{c}$
Now that's just the barest of sketches, I know, but hopefully it helps a bit. Again, I strongly recommend that you get ahold of the Scott piece if you want to see all the details worked out.