Suppose $R$ is a commutative ring, $I\subseteq R$ a principal ideal, and we're given split short exact sequences \begin{align*} R \to R^n &\to R^{n-1} \\ R/I \to (R/I)^n &\to (R/I)^{n-1} \end{align*} the first inducing the second on quotients. Suppose further that there is another splitting of the middle terms \begin{align*} R^n &\cong S \times T \\ (R/I)^n &\cong S_I \times T_I \end{align*} (again the former inducing the latter on quotients) consisting of free $R$-modules ($R/I$-modules) $S= R^{n-1}$ and $T= R$ (resp. $S_I =(R/I)^{n-1}$ and $T_I=R/I$).

A priori the direct summand $R$ specified in the first s.e.s. can embed into $S$, or into $T$ or diagonally into both. My question concerns its image under the assumption that on quotients the rank one copy $R/I$ embeds as direct summand into $S_I$.

I can see that this eliminates the possibility that $R$ embeds only into $T$ and restricts a diagonal embedding to the situation where the projection of $R$ to $T$ has to be contained in the ideal $I$.

Question: Is it possible that to show that, in case of a diagonal embedding, the projection of $R$ to $S$ is a direct summand of $S$?