Let $C_K$ be the congruence defined by an ideal $K$.

Then, we have three maps defined on $S = C_I \otimes_R C_J$:

   * $\pi_0 : S \to C_I$ induced by the projection $C_J \to R$
   * $\pi_1 : S \to C_J$ induced by the projection $C_I \to R$
   * $\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$.

Let $T = \{ x \in S \mid \pi_0(x) \in R \wedge \pi_1(x) \in R \}$. I claim that $\mu(T) = C_{IJ}$.


On the level of $R$ modules, we have the evident isomorphism $C_K \cong R \oplus K$, and we have

$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$

The maps $\pi_i$ are projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with.

By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form
$$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$
and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.