Quantify how small are the integrals $\int_0^N e^{-Nx}\binom{x}{N}dx $, as $N\to\infty$

I did some calculations and it seems like your integral, call it $f(N)$, is approximately $-(-1)^N/N^3$ as $N$ gets large. You may be able to get more accurate approximation.

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

$$I_N = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2\log N}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right). $$