When working with tensor products it is often easier to define homomorphisms in both directions and show that they are inverse to each other, rather than showing surjectivity and injectivity for one of them. In this case, you can define
\begin{align*}
A &\longrightarrow R/I \otimes_R A,\\
a &\longmapsto \overline{1} \otimes a
\end{align*}
whose kernel contains $IA$ (since $\overline{1} \otimes ia = \overline{i} \otimes a = 0\otimes a = 0$), so induces a well-defined homomorphism
\begin{align*}
A/IA &\longrightarrow R/I \otimes_R A,\\
\overline{a} &\longmapsto \overline{1} \otimes a.
\end{align*}
Now show that this is inverse to the map $R/I \otimes_R A \to A/IA$ that you already defined. By the way, all homomorphisms here are homomorphisms of $R$-modules, not only of abelian groups.