Let me write a bit of what seems to me that natural/naive approach, and perhaps the questioner can comment on the direction...

From $\int_0^\infty y^s\,e^{-ty}\;{dy\over y}=t^{-s}\,\Gamma(s)$, writing $Z(s)$ for your zeta, 
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; \int_0^\infty y^{s/2}\,\sum_{n\in \mathbb Z} e^{-\pi (n^2+c)y}\;{dy\over y}
$$ 
Unlike Riemann's argument, the presence of $c>0$ will cause Poisson summation to produce an expression that behaves well, without breaking the integral into two pieces. Namely, the Fourier transform of $x\to e^{-\pi (x^2+c)y}$ is ${1\over \sqrt{y}} e^{-\pi (x^2/y +cy)}$, so this becomes
$$
\sum_{n\in \mathbb Z} \int_0^\infty y^{{s-1\over 2}} e^{-\pi (n^2/y+cy)}\;{dy\over y}
$$
The $n=0$ term should be taken out, and is elementary. For the rest, replace $y$ by $\sqrt{c} y/n$, to get
$$
\pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s})
\;=\; (n=0) + \sum_{n\not=0}(\sqrt{c}/n)^{(s-1)/2}  \int_0^\infty y^{{s-1}\over 2} e^{-\pi \sqrt{c}\,n\,(y+1/y)}\;{dy\over y}
$$
The right-hand side seems to be entire in $s$, so the factor of $\Gamma(s/2)$ on the left-hand side means that the factor $Z(s)-c^{-s}$ vanishes at $s=0$. Thus, the right-hand side is close to computing the derivative of $Z(s)-c^{-s}$ at $s=0$.

But/and this makes me wonder whether about that extra term $c^{-s}$. It is needed to make Poisson summation work, but then it doesn't seem to me that $Z(s)=0$, so $Z'(0)$ would not be the leading term, which I would have thought would have been the object of interest.

In any case, the right-hand side is a sum of values of Bessel functions, but not obviously (to me) further simplifiable at $s=0$. At $s=1$, the sum over $n$ admits summing as geometric series, so at least the outcome is just an integral  in $y$... though it seems not elementary, still.

Comment?