This is but a partial answer.

If $c$ is integer, then you can add a few terms to reduce the problem to $\prod_{n=1}^\infty (1+an^b)$. If in addition $b$ is integer, then you can write $$(1+an^b) = \prod_{\xi^b=-a}(1+(-\frac \xi n))$$ and reduce to a finite product of inverses of $\Gamma$ functions and exponentials, using the product formula for $\Gamma$ (at complex values of the parameter).

I don't know what to make of the general case, but there are more knowledgeable people out here.

This is but a partial answer.

If $c$ is integer, then you can add a few terms to reduce the problem to $\prod_{n=1}^\infty (1+an^b)$. If in addition $b$ is integer, then you can write $$(1+an^b) = \prod_{\xi^b=-a}(1+(-\frac \xi n))$$ and reduce to a finite product of inverses of $\Gamma$ functions and exponentials, using the product formula for $\Gamma$ (at complex values of the parameter).

I don't know what to make of the general case, but there are more knowledgeable people here.