Show that $\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}$

You're just one step away: for $b>0$, we have $$\int_0^\infty {\frac{1}{{{x^\alpha }(b + x)}}dx} = {b^{ - \alpha }}\int_0^\infty {\frac{{{x^{ - \alpha }}}}{{1 + x}}dx} = \frac{{{b^{ - \alpha }}\pi }}{{\sin \alpha \pi }}$$ so $$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}dx=-\frac{1}{\Gamma(s)}\int_0^{\infty}\int_0^{\infty}\frac{t^{s-1}}{x^{\alpha}(e^t+x)}dxdt = - \frac{\pi }{{\Gamma (s)\sin \alpha \pi }}\int_0^\infty {{t^{s - 1}}{e^{ - \alpha t}}dt} $$

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The series expansion of $\text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.

I have finally figured out how to use Ramanujans Master Theorem in this case.

Ramanujans Master Theorem

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)\mathrm dx=\Gamma(s)\phi(-s)$$

Lets get back to the given integral. The series representation of $\operatorname{Li}_s(-x)$ is given by $\displaystyle\sum_{k=1}^{\infty}\frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $\displaystyle\phi(k)=\frac{\Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-\alpha$ yields to

$$\int_0^{\infty}x^{-\alpha-1}\operatorname{Li}_s(-x)\mathrm dx=\Gamma(-\alpha)\phi(\alpha)=\Gamma(-\alpha)\frac{\Gamma(\alpha+1)}{\alpha^s}$$

This term can be simplified by using Eulers Reflection Formula with $z=\alpha+1$ which finally leads to

$$\frac1{\alpha^s}\Gamma(\alpha+1)\Gamma(-\alpha)=\frac1{\alpha^s}\frac{\pi}{\sin(\pi(\alpha+1))}=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi\alpha)}$$