Are all commutative, associative binary operations isomorphic to addition?
You can consider an operation that does not give the structure of a group. For instance $$x\circ y\ \colon = \max(x,y)$$ Here we have $x \circ x = x$. In fact we have a lattice.
But even if the operation would give a commutative group, it may not be isomorphic to $\mathbb{R}$ with addition. For instance $\mathbb{R}/\mathbb{Z}$ is one such group of the same cardinality as $\mathbb{R}$.
You can check that an abelian group with the same cardinatlity as $\mathbb{R}$ is isomorphic to $\mathbb{R}$ with addition if an only if it is a uniquely-divisible group: for every $x$ and $n\ge 1$ integer, there exist a unique $y$ such that $ny = x$. Since then it is a vector space over $\mathbb{Q}$. However, a concrete isomorphism may not be available without some axiom of choice things. For instance, $\mathbb{R}/\mathbb{Q}$ with addition.
Edit: The new version of your question is, I think, much more general than you intended; the answer is yes, due to the following cheap trick, but it doesn't really have anything to do with addition anymore.
Let $m : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be any commutative operation (we will not need associativity, and operations of the form $g(f(x) + f(y))$ will not be associative in general). We want to find $g, f$ such that $m(x, y) = g(f(x) + f(y))$. You say that $f$ can take values in "some field." We'll take this field to be the function field $K = \mathbb{Q}(f_x, x \in \mathbb{R})$ over $\mathbb{Q}$ on uncountably many variables, one for every real number, and $f : \mathbb{R} \to K$ will be the function $f(x) = f_x$. This guarantees that $f(x) + f(y) = f(z) + f(w)$ iff either $x = z$ and $y = w$ or $x = w$ and $y = z$; in other words, the value of $f(x) + f(y)$ uniquely determines the unordered pair $\{ x, y \}$.
Now we can take $g : K \to \mathbb{R}$ to be any function such that $g(f_x + f_y) = m(x, y)$; since $f_x + f_y$ uniquely determines $\{ x, y \}$ and $m$ is commutative this is well-defined. Of course we have not used any features of $\mathbb{R}$ in this argument, which can be replaced by an arbitrary set.
(This responds to the original question only.)
Your statement about addition and multiplication is not quite right; you only get an isomorphism between $(\mathbb{R}, +)$ and $(\mathbb{R}_{+}, \times)$ (where the subscript denotes taking positive reals). However, it is true that addition and multiplication are locally isomorphic in a suitable sense.
Are all commutative, associative operations on the reals isomorphic to addition in a similar way?
As stated, since you haven't asked for compatibility with any other structure on the reals, you are just asking for commutative monoids with the same cardinality as the reals. There is an enormous diversity of such things, even if we restrict further to abelian groups (which means requiring inverses); as a random example, any countable product $\prod_{i=1}^{\infty} A_i$ of at-most-countable but nontrivial abelian groups has the same cardinality as the reals, for example $\prod_{i=1}^{\infty} \mathbb{Z}$, but such a group is isomorphic to the reals as an abstract group iff each $A_i$ is an at-most-countable-dimensional $\mathbb{Q}$-vector space.
However, we can get a positive result by requiring continuity.
Proposition: Every continuous abelian group structure on $\mathbb{R}$ (with its usual topology) is continuously isomorphic to addition.
This came up recently on MathOverflow. It follows from Gleason-Montgomery-Zippin's solution to Hilbert's fifth problem, which is almost certainly overkill; probably it follows from an easier and known structure theorem for locally compact (Hausdorff) abelian groups but I couldn't quite finish the argument. Maybe it can also be done "by hand."
(Edit: This is overkill, as expected. There is a totally elementary argument that a topological group (not necessarily abelian!) homeomorphic to $\mathbb{R}$ must be isomorphic to $(\mathbb{R}, +)$; see this math.SE question.)
If you further require smoothness then the proof is easier and can probably be done "by hand" more easily; abstractly, $\mathbb{R}$ is the unique $1$-dimensional Lie algebra and hence the unique $1$-dimensional simply connected Lie group, so:
Corollary: Any smooth abelian group structure on $\mathbb{R}$ is smoothly isomorphic to addition.
A related result you may be interested in is the classification of $1$-dimensional formal group laws; over a $\mathbb{Q}$-algebra, and in particular over $\mathbb{R}$, they are all isomorphic to the additive formal group law.
A question I don't know the answer to is whether it suffices to require measurability; that is, I don't know whether a measurable abelian group structure on $\mathbb{R}$ (with the Borel $\sigma$-algebra, say) is measurably isomorphic to addition.