According to nlab, the Grothendieck group $K(\mathfrak{A})$ of a small Abelian category $\mathfrak{A}$ is the set of isomorphism classes of objects in $\mathfrak{A}$, and $[A]+[B] = [C]$ iff there is a short exact sequence $0\to A\to C\to B\to 0$.
Clearly $[A]+[B] = [A\oplus B]$ since $0\to A\to A\oplus B\to B\to 0$ is short exact.
To show that addition is well-defined, we have to show that for every short exact sequence $0\to A\to C\to B\to 0$, there exists an arrow $A\oplus B\to C$ making the obvious diagram commute, (Five Lemma implies that any such arrow is an isomorphism), but I can't figure out why such an arrow exists?
An abstract proof (one without assuming $\mathfrak{A}$ as sets) would be greatly appreciated, but I acknowledge we can argue with R-modules and use the embedding theorem.
The statement you are asking for is not true, but also not necessary. Note that there is an error in your first line: the Grothendieck group is not the set of isomorphism classes in $\mathfrak{A}$, but is only generated by these, modulo the relations from short exact sequences. Thus whenever
$$0\to A\to C\to B\to 0$$
is a short exact sequence in $\mathfrak{A}$, it is indeed the case that $[C]=[A\oplus B]$ in $K(\mathfrak{A})$. This does not imply or require that $C\cong A\oplus B$, it just means that $[C]$ turns out to be a redundant generator, because it is equal to the generator $[A\oplus B]$. This behaviour is even generic: most abelian categories of interest (e.g. modules over a non-semi-simple ring) have short exact sequences in which the middle term is not isomorphic to the direct sum of the outer ones. If a Jordan–Hölder theorem holds in $\mathfrak{A}$, then $[X]\in K(\mathfrak{A})$ remembers only the multiplicities of the simple composition factors of $X$, not the isomorphism class.