Expansion of the Mittag-Leffler function $E_{s}(x^{s})$

Question

I want to find an expansion of the Mittag-Leffler function $E_{s}(x^{s})$ of the form : $$E_{s}(x^{s})=\sum_{n=0}^{\infty}f_{n}(x)g_{n}(s)\;\;\;\;\;(\Im(x)=\Im(s)=0)$$ I first defind the function $\Omega(x,s)$: $$\Omega(x,s)=sE_{s}(x^{s})-s-(e^{x}-1)=\sum_{n=1}^{\infty}\frac{x^{ns}}{n\Gamma(ns)}-\frac{x^{n}}{n!}$$ The reason for the introduction of the term $e^{x}-1$ is to extend the domain of convergence of the Laplace transform of $\Omega(x,s)$, which is given by : $$\int_{0}^{\infty}\Omega(x,s)e^{-zx}dx=\frac{s}{z(z^{s}-1)}-\frac{1}{z(z-1)}$$ Applying the Bromwich inversion formula, we obtain : $$\Omega(x,s)=\frac{1}{2\pi i }\int_{c-i\infty}^{c+i\infty}\left[\frac{s}{z(z^{s}-1)}-\frac{1}{z(z-1)} \right ]e^{zx}dz\;\;\;\;(c>0)$$ Making the change of variables $z=e^{y}$, we obtain : $$\Omega(x,s)=\frac{1}{2\pi i }\int_{\Gamma}\left(\frac{s}{e^{sy}-1}-\frac{1}{e^{y}-1} \right )e^{xe^{y}}dy$$ Where the path $\Gamma$ is the map from $\Re(z)=c\;$ by the log function. Using the fact that : $$e^{x(e^{y}-1)}=\sum_{n=0}^{\infty}\frac{\phi_{n}(x)}{n!}y^{n}$$ $\phi_{n}(x)$ being the nth order Bell polynomial We obtain : $$\Omega(x,s)=\frac{1}{2\pi i }\sum_{n=0}^{\infty}\frac{e^{x}\phi_{n}(x)}{n!}\int_{\Gamma}\left(\frac{s}{e^{sy}-1}-\frac{1}{e^{y}-1} \right )y^{n}dy $$ Now, i am a bit confused on how to do the integrals above. Any help is highly appreciated.