First off, there should $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question). 

The identity can be derived from [my answer](https://mathoverflow.net/q/395897) to the previous question, which essentially represents the case of $R=1$.

We will also need the case $R=0$, for which we have
\begin{split}
&\,\sum_{i=0}^{2k} \big( \binom{n-1}i + (-1)^{i+1} \binom{n+i}i\big) i! S(2k+1,i+1) \\
=&\, \sum_{i=0}^{2k} \big( (n-1)_i - (-n-1)_i \big) S(2k+1,i+1) \\
= &\, n^{2k} - (-n)^{2k} = 0.
\end{split}

Now, having the identity established for $R=0$ and $R=1$, for larger $R$ it follows by induction thanks to Pascal's rule.