Asymptotic formula for $\sum_{k=N}^\infty \frac{x^k}{k!}$ as $N \to \infty$ This seems like a weird question, because the series has good convergence and there's no need to use other methods to estimate it for $N \to \infty$.
However, after seeing this question, I tried to come up with some approximation which could allow us to treat integrals containing this sum in the denominator.
Regardless of the particular application, I just wanted to ask if there's a way to approximate this sum using a finite combination of elementary or special functions?
Not counting the obvious $e^x-\sum_{k=0}^{N-1} \frac{x^k}{k!}$ of course.
Euler-Maclaurin formula doesn't seem very promising, because the resulting integral is even more complicated $$\int_N^\infty \frac{x^y}{\Gamma(y+1)}dy$$ And the derivatives containing various combinations of polygamma functions are also a pain to write. 
If there's no better asymptotic expression than the sum itself, so be it.

In another way, note that your sum is related with the Regularized Incomplete Gamma Function.
$$
\eqalign{
  & Q(s,z) = {{\Gamma (s,z)} \over {\Gamma (s)}} = e^{-z} \sum_{k = 0}^{s} {{{z^{k}} \over {k!}}}
  = \left( {1 - e^{-z} \sum_{k=s}^{\infty } {{{z^k} \over {k!}}} } \right)\quad  \Longrightarrow   \cr 
  &  \Longrightarrow \quad \sum _{k=s}^{\infty } {{{z^{k}} \over {k!}}}  = e^{z} P(s,z) = e^{z} {{\gamma (s,z)} \over {\Gamma (s)}} \cr} 
$$
note that here the summation is taken in the Antidelta meaning.
Therefore, using the asymptotic expansion of $Q(s,z)$ for $s \, \to \, \infty$ we get
$$
\sum_{k=s}^{\infty } {{{z^k} \over {k!}}}  = \sum_{k = s}^{\infty } {{{z^k} \over {k!}}}
  = e^{z} P(s,z) = {{s^{- s - 1/2} e^{s} z^{s} } \over {\sqrt {2\pi } }}\left( {1 + {{12z - 1} \over {12s}}
 + O\!\left( {{1 \over {s^{\,2} }}} \right)} \right),\quad  \left| s \right| \to \infty.
$$
