Let $\theta>0$ be irrational, and define
$$\zeta_\theta(s) = \sum_{n=1}^\infty \frac{\lfloor n \theta \rfloor}{n^s}.$$
This converges to an analytic function on $\Re(s)>2$. Does $\zeta_\theta(s)$ have a meromorphic continuation to $\mathbb{C}$?

Some motivation: I'm interested in zeta functions associated to distance functions on $\mathbb{R}^n$: For a distance function $d$, consider 
$$\zeta_d(s)=\sum_{\mathbf{m}\in \mathbb{Z}^n}' d(\mathbf{m})^{-s}.$$ 
If $d$ is smooth outside $0$, one can show $\zeta_d$ has a meromorphic continuation. If $d(x,y)=\max\{|y|,|x|/\theta\}$, then $\zeta_\theta(s)$ is $d(x,y)^{-s}$ summed over the integral points in the sector $\{(x,y)\ |\ y\theta>x>0\}$. Meromorphic continuation of $\zeta_\theta$ and $\zeta_{\theta^{-1}}$ would then imply the meromorphic continuation of $\zeta_d$.

Based on numerical evidence, I conjecture that for $\theta$ a quadratic irrational, $\zeta_\theta(s)$ has a meromorphic continuation to (at least) a region $\Re(s)>\sigma_0$, where $\sigma_0<0$, and that its poles in $\Re(s)\geq 0$ occur at $s=2,1$ and a subset of $it_0\mathbb{Z}$ for some $t_0\in \mathbb{R}$. A more tentative conjecture is that this is true for all real algebraic numbers.