What is this series relating to the residues of the Gamma function?

It is $f(s) = \gamma(s,1)$ the lower incomplete gamma function. You have separated $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$ into $$\Gamma(s) = \int_0^1 x^{s-1} e^{-x}dx+\int_1^\infty x^{s-1} e^{-x}dx$$ where $\int_1^\infty x^{s-1} e^{-x}dx$ is entire and $\int_0^1 x^{s-1} e^{-x}dx=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{s+n-1}dx$ is a locally uniformly convergent series of poles

(using say the Riemann-Lebesgue lemma and $\Gamma(s+1) = s \Gamma(s)$ we can see that $\gamma(s,1)\to 0$ when $s$ moves away from the negative real axis)

Let's compute

$$f(x+1)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(x+1+n)}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!(x+n)}$$

and observe that

$$f(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n-1}[(x+n)-x]}{n!(x+n)}=1-\frac 1e+x\sum_{n=1}^\infty\frac{(-1)^n}{n!(x+n)}=1-\frac 1e+x\left(f(x)-\frac 1x\right)$$

We get :

$$\boxed{\forall x>0,\,f(x+1)=x\,f(x)-\frac 1e}$$

This is not the same functional equation than the one verified by $\Gamma$, but looks like ...

This series differs from $\Gamma(x)$ by an analytic function. For $\Re(x)>0$, $$ \Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \, dt \\ = \int_0^1 t^{x-1} e^{-t} \, dt + \int_1^{\infty} t^{x-1} e^{-t} \, dt \\ = f(x) + \Gamma(1,x), $$ by expanding the exponential as a power series and integrating term-by-term. $\Gamma(1,x)$ is the upper incomplete Gamma-function, and is an analytic function of $x$, while your series is the lower incomplete Gamma-function $\gamma(1,x)$. Meromorphic continuation implies that the equality derived above holds whenever $x$ is not a nonpositive integer: one can show that $f(x)$ is locally uniformly convergent on domains avoiding the nonpositive integers, so is a meromorphic function, and the result follows.

Notably, it is easy to check using integration by parts that $\Gamma(1,x)$ is exponentially small for $x \ll 0$, hence why the dominant behaviour on the left is captured by $f$. On the other hand, $f$ is much smaller than the exponentially large $\Gamma(1,x)$ for $x \gg 0$, so $\Gamma(1,x)$ is dominant on the right.

For more information, you can also see this answer I wrote a while ago.