Your augmentation solution looks like it should work.

The maps $\text{id}: R \oplus I \to R \oplus I$ annd $R \oplus J \to R \oplus I : (r,j) \mapsto (r,0)$ induce

$$ R \oplus I \oplus J \oplus (I \otimes J)  \to R \oplus I : (r,i,j,k) \mapsto (r,i,0,0) $$

and similarly, for the maps to $R \oplus J$. The $I \otimes J$ summand is characterized as precisely the subgroup of things that vanish under both maps. The subring $R \oplus (I \otimes J)$ is precisely the elements whose image under both maps lies in $R$.