[EDIT]: After getting a nice counter example provided by Steven Landsburg I realize that I forgot to impose an important condition...namely $R$ is supposed to be complete w.r.t. the $I$-adic topology. (In particular, this implies that elements of the form $1+i$ for $i\in I$ are units.)

Suppose $R$ is a commutative ring complete w.r.t. the $I$-adic topology, where $I\subseteq R$ a principal ideal generated by a non-zero divisor. Assume 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 $S$?