The Mellin transform for:
$$\displaystyle \Gamma(s)\,\zeta(s)= \int_0^\infty x^{s-1} \frac{1}{e^x-1}\,dx$$
equals:
$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\Gamma(x)\,\zeta(x) \,dx =\frac{1}{e^s-1}$$
but is only valid for $c>1$, however if we take the Dirichlet $\eta$-function:
$$\displaystyle \Gamma(s)\,\eta(s)= \int_0^\infty x^{s-1} \frac{1}{e^x+1}\,dx$$
and transform it into:
$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\Gamma(x)\,\eta(x) \,dx =\frac{1}{e^s+1}$$
then it is valid for all $c >0$ and now allows for integrating over all complex values with $c=\frac12$.
Assuming the RH, this would mean that the integrand becomes zero for all $x=\rho=\frac12 \pm \gamma\,i$ or in other words: any value of $x$ that could encode information about the distribution of the primes is "knocked out" of the integral to be able to correctly produce $\frac{1}{e^s+1}$.
Wild question: suppose we could prove that there is absolutely no information about the distribution of primes in $\frac{1}{e^s+1}$, would that imply that all non-trivial zeros need to be on the critical line for the Mellin transform to be correct?
Just to share an additional learning I got based on the answer below:
For:
$$\displaystyle \dfrac{\Gamma(s)}{\zeta(s)}= \int_0^\infty x^{s-1} \sum_{n=1}^{\infty}\mu(n)\,e^{-n\,x}\,dx$$
the Mellin transform is ($\mu(n)$= the Mobius function):
$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\dfrac{\Gamma(x)}{\zeta(x)}\,dx =\sum_{n=1}^{\infty}\mu(n)\,e^{-n\,s}$$
and this immediately shows that the integrand would diverge when a non-trivial zero would lie off the critical line. Hence full convergence in $\frac12 < c < 1$ implies the RH.