Investigating the integral $\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1-x}\,dx \small \space\colon\space Re\{s\} \gt 1 \normalsize$,
With the special case ($s=3$) introduced here as $I_2=\int_{0}^{\infty}\frac{x^2}{e^x-1-x}\,dx$ :
$$ \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1-x} \,dx = \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1-x} \right) \,dx = \sum_{n=1}^{\infty} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{n} \,dx $$
Analyzing the last integral, it follows that:
$$
\begin{align}
& \small \quad \frac{0!}{\Gamma(s+0)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{1} \,dx = \zeta(s) \\
& \small \quad \frac{1!}{\Gamma(s+1)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{2} \,dx = \zeta(s) - \zeta(s+1) \\
& \small \quad \frac{2!}{\Gamma(s+2)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{3} \,dx = \zeta(s) - 3\zeta(s+1) + 2\zeta(s+2) \\
& \small \quad \frac{3!}{\Gamma(s+3)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{4} \,dx = \zeta(s) - 6\zeta(s+1) + 11\zeta(s+2) - 6\zeta(s+3) \\
& \small \quad \normalsize \text{. . .} \quad \text{. . .} \quad \text{. . .} \\
& \small \quad \frac{N!}{\Gamma(s+N)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{N+1} \,dx = \sum_{n=0}^{N} (-1)^{n}\,a_{n}\,\zeta(s+n) \\
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
Where the coefficients $a_n$ calculated from the product expansion:
$$ \prod_{n=1}^{N} \left( 1 - n s \right) = \sum_{n=0}^{N} (-1)^{n}\,a_{n}\,s^{n} = 1 - a_1\,s^1 + a_2\,s^2 - a_3\,s^3 + \text{ ... } \pm\,a_{\small N \normalsize}\,s^{\small N \normalsize}\\
\boxed{ \quad \Rightarrow \sum_{n=0}^{\infty} (-1)^{n}\,a_{n}\,\zeta(s+n) = \lim_{N\rightarrow\infty} \frac{N!}{\Gamma(s+N)} \int_{0}^{\infty} x^{s-2} \left( \frac{x}{e^x-1} \right)^{N+1} \,dx \quad } $$
And the question is about the last limit. In other words, does the limit exist and equal zero, so: $$\sum_{n=0}^{\infty}(-1)^{n}\,a_{n}\,\zeta(s+n) = 0 \qquad\colon\space Re\{s\} \gt 1 \tag{1}$$
On the other hand, we know that (except for $s=0$) the infinit product is not convergent:
$$ \prod_{n=1}^{\infty} \left( 1 - n s \right) = 1 + \sum_{n=1}^{\infty} (-1)^{n}\,a_{n}\,s^{n} = 1 - a_1\,s^1 + a_2\,s^2 - a_3\,s^3 + \text{...} $$
Is it possible to utilize the exceptional case using the limit $\left\{\small\lim_{z\rightarrow\infty}\zeta(z)=1\normalsize\right\}$ to show that: $$\sum_{n=0}^{\infty}(-1)^{n}\,a_{n}\left[\zeta(s+n)-1\right] = 0 \qquad\colon\space Re\{s\} \gt 1 \tag{2}$$