You can use the residue theorem, given the series expansion of $$h(z)=e^{b/(z-c)}g(z)=\sum_{n=0}^\infty h_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to $$\oint_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum_{n=1}^\infty \frac{ h_{n-1}a^n}{n!}=2\pi i\sum_{n=1}^\infty \frac{ a^n}{n!(n-1)!}h^{(n-1)}(0),$$ with $h^{(n)}(0)$ the $n$-fold derivative of $h(z)$ evaluated at $z=0$. Whether or not this sum can be evaluated in closed form will depend on your choice for $g(z)$.

You can use the residue theorem, given the series expansion of $$h(z)=e^{b/(z-c)}g(z)=\sum_{n=0}^\infty h_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to $$\oint_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum_{n=1}^\infty \frac{ h_{n-1}a^n}{n!}=2\pi i\sum_{n=1}^\infty \frac{ a^n}{n!(n-1)!}h^{(n-1)}(0),$$ with $h^{(n)}(0)$ the $n$-fold derivative of $h(z)$ evaluated at $z=0$.

I would love to be shown wrong, but I'm pretty certain this is the best one can do in the general case $-$ there is no short-cut to the residue at an essential singularity.

You can use the residue theorem, given the series expansion $$e^{b/(z-c)}g(z)=\sum_{n=0}^\infty g_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to $$\oint_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum_{n=1}^\infty \frac{g_{n-1} a^n}{n!}.$$

You can use the residue theorem, given the series expansion of $$h(z)=e^{b/(z-c)}g(z)=\sum_{n=0}^\infty h_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to $$\oint_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum_{n=1}^\infty \frac{ h_{n-1}a^n}{n!}=2\pi i\sum_{n=1}^\infty \frac{ a^n}{n!(n-1)!}h^{(n-1)}(0),$$ with $h^{(n)}(0)$ the $n$-fold derivative of $h(z)$ evaluated at $z=0$. Whether or not this sum can be evaluated in closed form will depend on your choice for $g(z)$.