Taking into account that $\binom pq=0$ for nonnegative integers $q>p$, write 
$$S(n,m)=\frac{(n+m+1)!}{n!m!}T(n,m),$$
where
\begin{align*}
	T(n,m)&:=\sum_{l\ge0}\frac{1}{n+m-l+1}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\ 
	&=\sum_{l\ge0}\int_0^1 dx\,x^{n+m-l}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk \\ 
	&=\int_0^1 dx\,x^{n+m}\sum_{l\ge0}\sum_{\substack{j+k=l\\ j\ge0,k\ge0}}(-1)^{j}\binom nj\binom mk x^{-j} x^{-k} \\ 
	&=\int_0^1 dx\,x^{n+m}\sum_{j\ge0}\binom nj(-x^{-1})^j\;\sum_{k\ge0}\binom mk x^{-k} \\ 
	&=\int_0^1 dx\,x^{n+m}(1-x^{-1})^n(1+x^{-1})^m\ \\ 
	&=\int_0^1 dx\,(x-1)^n(1+x)^m\ \\ 
	&=(-1)^n\sum_{k=0}^m\binom mk\int_0^1 dx\,(1-x)^n x^k\ \\ 
	&=(-1)^n\sum_{k=0}^m\frac{m!}{k!(m-k)!}\frac{k!n!}{(k+n+1)!}\ \\ 
	&=(-1)^n\sum_{k=0}^m\frac{m!}{k!(m-k)!}\frac{k!n!}{(k+n+1)!}\ \\ 
	&=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{k=0}^m\binom{m+n+1}{m-k}\ \\ 
	&=(-1)^n \frac{m!n!}{(m+n+1)!} \sum_{j=0}^m\binom{m+n+1}j.  
\end{align*}
So,
\begin{equation}
	S(n,m)=(-1)^n\sum_{j=0}^m\binom{m+n+1}j, 
\end{equation}
which is simpler than your desired expression.