Here are some initial thoughts. Put $$ X(R)=\{e\in R: e^2=e \text{ and } er=re \text{ for all } r\in R\} $$ We can partially order this by declaring that $e\leq f$ iff $ef=e$. We then put $$ Y(R)=\{e\in X(R): 0\lt e \text{ and there is no } f\in X(R) \text{ with } 0 \lt f \lt e \} $$ One can check that $X(R)$ is a finite Boolean algebra under this order (with meet operation $e\wedge f=ef$ and join $e\vee f=e+f-ef$) so it is isomorphic to the lattice of subsets of its set of atoms, which is $Y(R)$. In particular, if $|Y(R)|=n$ then $|X(R)|=2^n$. For $e\in Y(R)$ we put [ $$ R[e] = Re = \{ x\in R : ex=xe=x\} ] $$ We can then define $p:R\to\prod_{e\in Y(R)}R[e]$ by $p(x)_e=ex$. It is standard that this is an isomorphism of rings.
Next, by hypothesis we have a bijection $f:R\to S$ that preserves multiplication. It follows that $f$ gives an isomorphism $X(R)\to X(S)$ of posets, and thus a bijection $Y(R)\to Y(S)$. As the sets $R[e]$ and the maps $p_e$ are defined using only the multiplicative structure, we see that $f$ gives an isomorphism $R[e]\to S[f(e)]$ of multiplicative monoids for each $e\in Y(R)$. However, we do not obviously have an additive isomorphism from $R[e]$ to $S[f(e)]$, so this does not succeed in reducing the problem to the indecomposable case.
Nonetheless, it is worth thinking about the ring structure of $R[e]$. The quotient by the Jacobson radical is a finite simple ring and so is a matrix algebra over a finite division ring, but finite division rings are fields by a theorem of Wedderburn, so this quotient is quite tractable.
