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Taking into account that $\binom pq=0$ for nonnegative integers $p$ and $q$ such that $q>p$, 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\ \\ &=\int_0^1 dx\,(x-1)^n\,\sum_{k=0}^m\binom mk x^k\ \\ &=(-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 \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 (which seems to differ from the latter expression for $S(n,m)$ only by the sign).