[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 $S$?