Taking logarithm gives
$$
-\sum_{k=a}^N \log(e^{k \varkappa}-1)=
-\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa.
$$
The last sum is a Riemann sum of the integral
$$
\int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx=
\text{Li}_2(e^{\varkappa a})-
\text{Li}_2(e^{\varkappa N}),
$$
where $\text{Li}_2(x)$ is a [polylogarithm function][1].
So the product is approximately equal to
$$
e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})-
\text{Li}_2(e^{\varkappa a})}.
$$

  [1]: http://en.wikipedia.org/wiki/Polylogarithm