$\frac{1}{e^x-1}$, $\Gamma(s)$, $\zeta(s)$, and $x^{s-1}$

Just to give a few examples, we have that

$$\eqalign{ & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}dx} = \Gamma \left( s \right)\zeta \left( s \right) \cr & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} + 1}}dx} = \left( {1 - {2^{1 - s}}} \right)\Gamma \left( s \right)\zeta \left( s \right) =\eta(s)\Gamma(s) \cr & \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x}\left( {{e^x} - 1} \right)}}dx} = \Gamma \left( s \right)\left( {\zeta \left( s \right) - 1} \right) \cr & \int\limits_0^\infty {\frac{{{x^{s-1}}{e^x}}}{{{{\left( {{e^x} - 1} \right)}^2}}}dx} = \Gamma \left( {s } \right)\zeta \left( s-1 \right)\cr & \int\limits_0^\infty  {\frac{{{e^x}{x^{s - 1}}}}{{{{\left( {1 + {e^x}} \right)}^2}}}}  = \Gamma \left( s \right)\eta \left( s-1 \right) \cr} $$

Is there any theory that enables us to state that any integral of the form

$$\int\limits_0^\infty {F\left( {\frac{1}{{{e^x} - 1}},{e^x},{x^s}} \right)dx} $$

will necessarily be evaluated in terms of $\zeta$ and $\Gamma$?

I can offer the following observation. Consider the following identities \begin{align} \frac{1}{e^{x} - 1} = \sum_{k \geqslant 0} e^{-kx}, \quad \frac{1}{e^{x} + 1} = \sum_{k \geqslant 1} (-1)^{k + 1} e^{-kx} \quad \text{and} \quad \int_{0}^{\infty} x^{s} e^{kx} \ d^{\times} x = \Gamma(s) \, k^{-s}, \end{align} where $d^{\times} x = \frac{dx}{x}$ is the Haar invariant measure on $\mathbb{R}$. For $\mathsf{Re}(s) > 1$, $\mathsf{Re}(a) < 1$ and $a \not \in \mathbb{Z}$, one has \begin{align} \int_{0}^{\infty} \frac{x^{s} e^{a x}}{e^{x} - 1} \ d^{\times} x & = \int_{0}^{\infty} \sum_{k \geqslant 1} x^{s} e^{(a - k) x} \ d^{\times} x \\ & = \Gamma(s) \sum_{k \geqslant 1} (k - a)^{-s} \\ & = \Gamma(s) \zeta(s,1-a), \end{align} where $\zeta(s,1-a)$ denotes Hurwitz zeta function. Similarly, one has \begin{align} \int_{0}^{\infty} \frac{x^{s} e^{a x}}{e^{x} + 1} \ d^{\times} x & = \int_{0}^{\infty} \sum_{k \geqslant 1} (-1)^{k+1} x^{s} e^{(a - k) x} \ d^{\times} x \\ & = \Gamma(s) \sum_{k \geqslant 1} (-1)^{k+1} (k - a)^{-s} \\ & = \Gamma(s) 2^{-s} \left( \zeta(s,\tfrac{1-a}{2}) - \zeta(s,1 - \tfrac{a}{2}) \right). \end{align} The interchange of the summations and the integral sign is allowable since the summations are absolutely and uniformly convergent on their domains. These two formulas generalize the first two of your quoted formulas. The third one follows by splitting the integral into two using partial fractions and the integral formula of the Gamma function.