Path From Positive Dedekind Cuts to Reals?

This falls out from the the definition of a field and the requirement that $P$ embeds in the "group of differences" (your $P^2/\sim$). It will amount to saying that $P$ must be a semiring whose additive semigroup satisfies the cancellation law, if $x + y = x + z$, then $y = z$, and whose multiplicative semigroup is a group.

By the way, did you know that any densely and completely ordered commutative group is isomorphic to the reals? This leads to a little-known but very neat way of defining multiplication by studying the order-preserving homomorphisms of the additive group. I find this a far more satisfactory way of avoiding all those case splits.

Rather that define dedekind cuts on the set of rational numbers, in a natural way we can modify the definition and define cuts on the positive rational numbers. In this new setting, we construct the non-negative real numbers, $\mathbb R^{\ge0}$, and we can show it forms a commutative semifield, and we don't get bogged down in checking, ad nauseum, the case-by-case breakdown for multiplication.

In exactly the same way as we the integers are constructed from the natural numbers found here, we can get the negative real numbers from $\mathbb R^{\ge0}$, giving us the set of all real numbers. Using the $\mathbb N_0 \to \mathbb Z$ theory, it will be a ring.

Proposition: Every nonzero real number $x$ has a multiplicative inverse.
Proof
If the pair $(u,v)$ with $u,v \in \mathbb R^{\ge0}$ represents $x$, then $u \ne v$.
Recall the definition of multiplication:

$\quad {\displaystyle [(a,b)]\cdot [(c,d)]:=[(ac+bd,ad+bc)]}$

If $u \lt v$, then the number (block) represented by $(0, \frac{1}{v-u})$ is the inverse of $x$.

If $u \gt v$, then the number (block) represented by $(\frac{1}{u-v},0)$ is the inverse of $x$.

So every nonzero element of $\mathbb R$ has a multiplicative inverse and so we have actually constructed a field of numbers. $\quad \blacksquare$