Let's define : $f(t)=\sum_{0}^{\infty}\tfrac{a^n}{(n!)(c-bn)}e^{(c-bn)t}$
then $f'(t)=\sum_{0}^{\infty}\tfrac{a^n}{n!}e^{(c-bn)t}=e^{ct}\sum_{0}^{\infty}\tfrac{(a e^{-bt})^n}{n!}=e^{ct+ae^{-bt}}$
EDIT (the $t$ was missing in $ct$ !)
for $u=ae^{-bt}$ that is $t=-\frac{\log(\frac ua)}b$ we have : $\int e^{ct+ae^{-bt}} dt= -\frac 1b \int \frac{e^{-\frac{c\log(\frac ua)}b}e^{u}}u du=-\frac 1b \int \left(\frac ua\right)^{-\frac cb}\frac{e^{u}}u du$ $=-\frac {a^{\frac cb}}b \int u^{-1-\frac cb}e^{u} du=-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-u\right)$
with $\gamma$ the 'lower incomplete gamma function' getting : $f(t)=C(a,b,c)-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-ae^{-bt}\right)$
where $C(a,b,c)=0$ I think.
This is a clearly non elementary result that we may rewrite as : $f(t)=\frac {(-a)^{\frac cb}}b\left[\Gamma\left(-\frac cb,-ae^{-bt}\right)-\Gamma\left(-\frac cb\right)\right]$
The same result was obtained by Alpha ('Alternate form' assuming $a,b,c,t$ positive) :
