I've tried to do some computations. Although I did not obtain the required formula, I am posting it here - hopefully someone can either spot a mistake in what I did or modify the method to obtain the above result.
$\sum\limits_{i=0}^k \sum\limits_{i=0}^k \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^k \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}+ \sum\limits_{m=k+1}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!} $
The first sum:
$\sum\limits_{m=0}^k \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^k \sum\limits_{i=0}^m \frac{x^iy^{m-i}}{i!(m-i)!}= \sum\limits_{m=0}^k \frac1{m!} \sum\limits_{i=0}^m \binom mi x^iy^{m-i}= \sum\limits_{m=0}^k \frac{(x+y)^m}{m!} $
The second sum:
$\sum\limits_{m=k+1}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{n=1}^k \sum\limits_{\substack{i+j=k+n\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{n=1}^k \sum\limits_{i=n}^k \frac{x^i}{i!} \frac{y^{n+k-i}}{(n+k-i)!} $
The difference $e^xe^y-e^{x+y}$ is the limit of the last sum.