Here is a confirmation of @Somos' computation. Notice that
$$ \binom{x}{N} = \frac{x(x-1)\cdots(x-N+1)}{N!} = \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] x^k, $$
where $\left[ {N \atop k} \right]$ is the unsigned Stirling numbers of the first kind. Plugging this back and computing,
\begin{align*}
I_N := \int_{0}^{N} e^{-Nx} \binom{x}{N} \, \mathrm{d}x
&= \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \int_{0}^{N} x^k e^{-Nx} \, \mathrm{d}x \\
&= \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}}(1 - \epsilon_{N,k}),
\end{align*}
where $\epsilon_{N,k} = \int_{N^2}^{\infty} \frac{x^k}{k!} e^{-x} \, dx $.
Estimation of error term. We first note that there exists a constant $C_1 > 0$ satisfying
$$ \epsilon_{N,k} \leq \epsilon_{N,N} \leq C_1 e^{-N}$$
for all $1 \leq k \leq N$. The first inequality is easily proved under certain probabilistic interpretation. Let $T_1, T_2, \cdots$ be independent random variables having exponential distributions. Then we can write $ \epsilon_{N,k} = \Bbb{P}(T_1 + \cdots + T_{k+1} > N^2 ) $. This proves that $\epsilon_{N,k}$ is monotone increasing in $k$. Next, apply the substitution $ x \mapsto x + N$ to write
$$ \epsilon_{N,N} = e^{-N} \int_{N^2 - N}^{\infty} \frac{(x+N)^N}{N!} e^{-x} \, dx. $$
Now notice that $x + N \leq \frac{N}{N-1} x$ for $x \geq N^2 - N$. Using this,
$$ \epsilon_{N,N} \leq
e^{-N} \left(\frac{N}{N-1}\right)^N \int_{0}^{\infty} \frac{x^N}{N!} e^{-x} \, dx
\leq C_1 e^{-N} $$
for some $C_1 > 0$. This bound is somewhat crude, but it is enough for our purpose. Next we recall the following identity
$$ \sum_{k=0}^{N} \left[ {N \atop k} \right] = N! $$
From this, we have
$$ \left| \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}} \epsilon_{N,k} \right|
\leq C_1e^{-N}. $$
Extracting the leading term. We remark the following identities: if $N \geq 1$, then
$$ \left[ {N \atop 0} \right] = 0, \qquad \left[ {N \atop 1} \right] = (N-1)!, \qquad \left[ {N \atop 2} \right] = (N-1)!N_{N-1}. $$
Since $k!/N^k$ is decreasing in $k$, for $k \geq 3$ we have $ k!/N^k \leq 6/N^3$. So
\begin{align*}
&\sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}} \\
&\hspace{2em} = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2H_{N-1}}{N^4} + \mathcal{O}\left( \sum_{k=3}^{N} \frac{1}{N!}\left[ {N \atop k} \right] \frac{6}{N^{4}} \right) \\
&\hspace{4em} = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2H_{N-1}}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right).
\end{align*}
Conclusion. Combining both estimates, we obtain
$$I_N = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2\log N}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right). $$
