As $R_R/N=\langle 1+N \rangle$, every $f\in End_R(R_R/N)$ is uniquely determined by $f(1+N)$. Then, one has only to prove that $f(1+N)\in S/N$, right?
If $f(1+N)\notin S/N$, then for $f(1+N)=r+N$ for a $r$ such that there is a $n\in N$ s.t. $rn\notin N$. But I cannot continue.
$S$ is a subring of $R$ and we have a map $\varphi\colon S\to\operatorname{End}_R(R/N)$, where $\varphi(r)=\hat{r}$ and $\hat{r}(x+N)=rx+N$.
The kernel of $\varphi$ is $N$: indeed, $\hat{r}=0$ if and only if $\hat{r}(x+N)=0+N$ for every $x\in R$, that is $rx\in N$ for every $x\in R$. In particular $r\in\ker\varphi$ implies $r\in N$. The converse is obvious.
Therefore $\varphi$ induces an injection $S/N\to\operatorname{End}_R(R/N)$. It remains to prove that $\varphi$ is surjective.
Take $f\in\operatorname{End}_R(R/N)$ and choose $r\in R$ such that $f(1+N)=r+N$. Then, for all $x\in R$, $f(x+N)=f((1+N)x)=f(1+N)x=(r+N)x=rx+N$. In particular $r\in S$ and $f=\hat{r}$.