[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 S \times T &\to R^{n-1} \\
R/I \to S\_I \times T\_I &\to (R/I)^{n-1}
\end{align\*}
the first inducing the second 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 $S$?
> 
> 
> 



Lifting a direct summand of a free module
Let $v$ be any map at all from $R$ to $S\approx R^{n-1}$.


Then $v=qu$ where $u:R\rightarrow R^n$ is given by $r\mapsto (vr,r)$ and $q:R^n\rightarrow S$ is the obvious projection. 


Therefore your map from $R\rightarrow S$ can be any map at all. You are therefore asking whether an arbitrary map that becomes a split injection mod $I$ must be a split injection. 


For a counterexample, let $n=1$, and let $v$ be multiplication by some non-unit $1+i$ with $i\in I$.


