It looks like the tentative conjecture is (almost) true -- $\zeta_\theta(s)$ extends to a meromorphic function on $\mathbb{C}$ for all real quadratic irrationals $\theta$ with poles at $s=2$, $s=1$ and in the infinite set of vertical lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{Z}_{\ge0}\rbrace$. I'm not sure about the more tentative conjecture for arbitrary algebraic numbers. In factAlso, more simply, it will extend to a meromorphic function over $\Re(s) > 0$ whenever $\theta$ cannot be approximated too well by rationals (in particular, if $\theta$ has irrationality measure 2 which, by the Thue-Siegel-Roth theorem, includes all algebraic irrationals). Conversely, if an irrational $\theta$ can be approximated too well by rationals then there will not be a meromorphic extension, which will be the case outside of a meagre subset of $\mathbb{R}^+$$\mathbb{R}$.
[I think[Edit: As Pieter pointed out in the following is allcomments, my initial answer carelessely missed out a rather important term, so was not correct. I've updated the proof sketch to deal with this term, and restricted to quadratic irrationals for the meromorphic extension. This is very sketchy, and I only go through the full details for $\mathbb{Z}+\mathbb{Z}\theta$ a quadratic number field, but I would be more confident if someone double checksthink it.. does generalize to arbitrary quadratic irrationals in a similar way.]
First, let $\lbrace x\rbrace = x - \lfloor x\rfloor$ denote the fractional part of $x$, and write
$$
\tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{\lbrace n\theta\rbrace-1/2}{n^s},
$$
which converges, at least, on $\Re(s) > 1$ (at the very least).
Then
Then, the function $\zeta_\theta$ defined in the question is
$$
\zeta_\theta(s)=\theta\zeta(s-1)-\frac12\zeta(s)-\tilde\zeta_\theta(s),
$$
where $\zeta$ is the Riemann zeta function. So, at least, we have an extension to $\Re(s) > 1$. In the case where $\theta$ is rational then $\lbrace n\theta\rbrace-1/2$ is periodic in $n$, so $\tilde\zeta_\theta$ is a linear combination of Dirichlet L-functions and has a meromorphic extension to $\mathbb{C}$. Henceforth, I will only consider irrational $\theta$. Setting $F(x)=\sum_{1\le n\le x}(\lbrace n\theta\rbrace-1/2)$, the equidistribution theorem states that $F(x)/x\to0$ as $x\to\infty$. Rearranginging the expression for $\tilde\zeta_\theta$ gives
$$
\tilde\zeta_\theta(s)=s\sum_{n\ge1}\frac{F(n)}{n}n^{-s}\left(\frac ns(1-(1+1/n)^{-s})\right).
$$$$
\tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{F(n)}{n}n^{1-s}\left(1-(1+1/n)^{-s}\right)
$$
If $\lvert F(n)/n\rvert$ is bounded by some $\epsilon > 0$$\delta > 0$ then this expression will be bounded by $\epsilon\zeta(s)$$\delta\zeta(s)$ for all $\Re(s) > 1$$s > 1$. So, by equidistribution, it follows that $(s-1)\tilde\zeta_\theta(s)\to0$ as $s\to1$ (on $s > 1$). If a meromorphic extension existed in a neighbourhood of 1 then this would imply that $\tilde\zeta_\theta$ is bounded near 1. However, if $\theta=p/q$ is rational ($p,q$ coprime) then $F(x)/x\to-1/(2q)$ so, if $x$ is too closely approximated by rationals then there will be a slight bias in $F(x)/x$ and $\tilde\zeta_\theta(s)$ will not be bounded close to 1. I think that the existence of rational approximations $p_n/q_n$ with $q_n^2\log\lvert\theta-p_n/q_n\rvert\to-\infty$ is enough for this to happen.
Now, suppose that $\theta$ has irrationality measure 2no greater than some finite $\gamma$, so that there are only finitely many rational approximations $\lvert\theta-p/q\rvert\le q^{-2-\epsilon}$$\lvert\theta-p/q\rvert\le q^{-\gamma-\delta}$ for eachany $\epsilon > 0$$\delta > 0$. Equivalently, there are constantsis a constant $C_\epsilon > 0$$C_\delta > 0$ such that $n^{1+\epsilon}\lvert n\theta-m\rvert\ge C_\epsilon$$n^{\gamma-1+\delta}\lvert n\theta-m\rvert\ge C$ for all positive integer $n$ and integer $m$. Then, for each $\epsilon > 0$, the discrepancy of the set $\left\lbrace \lbrace k\theta\rbrace\colon 1\le k\le n\right\rbrace$ is bounded by $Kn^{-1+\epsilon}$$Kn^{-1/(\gamma-1)+\delta}$ (for a constant $K$, depending on $\epsilon > 0$). This means that $F(n)/n$ is bounded by $2Kn^{-1+\epsilon}$$2Kn^{-1/(\gamma-1)+\delta}$ and, hence, the expression above for $\tilde\zeta(s)$ converges for all$\tilde\zeta_\theta(s)$ has summand going to zero at rate $\Re(s) > \epsilon$$O(n^{-s-1/(\gamma-1)+\delta})$. So, $\tilde\zeta(s)$$\tilde\zeta_\theta(s)$ is an analytic function on $\Re(s) > 0$$\Re(s) > (\gamma-2)/(\gamma-1)$. ThereforeIn particular, for algebraic irrationals, $\zeta_\theta(s)$ is$\gamma=2$ and we get a meromorphic in this regionextension of $\zeta_\theta(s)$ to $\Re(s) > 0$ with simple poles at $s=2$ and $s=1$.
We canI now attempt to extend to the whole of $\mathbb{C}$ via a functional equation. Hurwitz's formula giveswhich, for $\Re(s) > 1$$\Re(s) > 0$ and irrational $x$ in the unit interval, gives
$$
\begin{array}{l}
&\sum_{n\ge1}e^{2\pi inx}n^{-s}=(2\pi)^{s-1}\Gamma(1-s)\left(e^{i\pi(1-s)/2}\zeta(1-s,x)+e^{i\pi(s-1)/2}\zeta(1-s,1-x)\right)
\end{array}
$$
where $\zeta(s,x)=\sum_{n\ge1}(n+x)^{-s}$$\zeta(s,x)=\sum_{n\ge0}(n+x)^{-s}$ is the Hurwitz Zeta function.
Plugging in the Fourier series $\lbrace x\rbrace=1/2+\sum_{k\not=0}(i/(2\pi k))e^{2\pi ikx}$, applying Hurwitz's formula, and being careless about when the sums converge/commute (it can be made more rigorous) gives,
$$
\begin{array}{rl}
\tilde\zeta_\theta(s)&=\sum_{n\ge1}\sum_{k\not=0}\frac{ie^{2\pi ikn\theta}}{2\pi kn^s}\cr
&=\sum_{k\not=0}\frac{i}{2\pi k}\sum_{n\ge1}\frac{e^{2\pi in\lbrace k\theta\rbrace}}{n^s}\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge1}\sum_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}.
\end{array}
$$
Setting
$$
\begin{array}{l}
&g_{n,x,s}=(1+\lbrace x\rbrace/n)^{-s}-(1+(1-\lbrace x\rbrace)/n)^{-s},\cr &G_{n,x,s}=\sum_{1\le k\le x}g_{n,k\theta,s},
\end{array}
$$$$
\begin{array}{rl}
\tilde\zeta_\theta(s)&=\sum_{n\ge1}\sum_{k\not=0}\frac{ie^{2\pi ink\theta}}{2\pi kn^s}\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge0}\sum_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}.
\end{array}
$$
we haveSplit up the summation as
$$
\begin{array}{rl}
\tilde\zeta_\theta(s)&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge1}\sum_{k\ge1}k^{-1}g_{n,k\theta,1-s}n^{s-1}\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge1}\sum_{k\ge1}\frac{G_{n,k,1-s}}{k(k+1)}n^{s-1}.
\end{array}
$$$$
\sum_{n\ge1}n^{s-1}\sum_{k\not=0}k^{-1}\left(1+\lbrace k\theta\rbrace/n\right)^{s-1}+\sum_{k\not=0}k^{-1}\lbrace k\theta\rbrace^{s-1}.
$$
NowThe first term can be handled easily. Use the fact that $\theta$ has irrationality measure 2, asso that $\left\lbrace\lbrace k\theta\rbrace\colon k=1,\ldots,n\right\rbrace$ has discrepancy $O(n^{-1+\delta})$. As the function $x\mapsto(1+x/n)^{s-1}$ has bounded variation of size $g_{n,x,s}$ on$O(1/n)$ over the unit interval, the summation over $0\le x\le1$$k$ converges and is bounded byof size $\lvert s\rvert/n$ for$O(1/n)$. Therefore, the sum over $\Re(s) > 0$$n$ converges on $\Re(s) < 1$, showing that the first term converges to an analytic function on $G_{n,k,s}$$\Re(s) < 1$.
The final summation over $k$ is of sizemore problematic. Assuming $O(\lvert s\rvert k^\epsilon /n)$$\theta$ is a quadratic irrational, and using $\bar z$ to denote the conjugate of $z\in\mathbb{Q}(\theta)$, henceit can be rewritten (up to order of summation) as
$$
(\bar\theta-\theta)\sum_{0 < z < 1}\frac{z^{s-1}}{\bar z-z}=(\bar\theta-\theta)\sum_{j=0}^\infty\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}.
$$
Here, I have substituted in $z=\lbrace k\theta\rbrace$, and the summation is over all $z\in\mathbb{Z}+\mathbb{Z}\theta$ lying in the unit interval. For each bounded region for $s$ on which we want to extend. We only actually have to expand out a finite number of terms in the sum over $\sum_{k\ge1}G_{n,k,s}/(k(k+1))$$j$ above, up until $\Re(s)+j-1$ is positive. In that case, the numerator of sizethe remainder is $O(\lvert s\rvert/n)$$z^{s+j-1}=\lbrace k\theta\rbrace^{s+j-1}$. ThereforeAs $x^{s+j-1}$ will have finite variation over the unit interval, the final expression above forremainder term will converge to an analytic function. So, we just need to show that each term in the sum over $\tilde\zeta_\theta(s)$$j$ extends to a meromorphic function.
I'll now restrict to the case where $\mathbb{Z}+\mathbb{Z}\theta$ is a sum overnumber field. For example, $n\ge1$ of$\theta=\sqrt{d}$ for squarefree $d\not\cong1$ (mod 4) or $\theta=(1+\sqrt{d})/2$ for squarefree $d\cong1$ (mod 4). I think the general quadratic case follows in a termsimilar manner, but it gets a bit messier. Let $\eta$ be the fundamental unit of size $O(\lvert 1-s\rvert n^{s-2})$$\mathbb{Z}[\theta]$ lying in the unit interval. Then, which convergeseach $z$ in the unit interval can be written uniquely as $\eta^ny$ for $y$ in $[\eta,1)$ and nonnegative integer $n$. Writing $\epsilon=\bar\eta\eta=\pm1$, the term for a fixed $j$ in the expansion above is (uniformly on compactsagain, being a bit careless about order of summation),
$$
\begin{array}{rl}
\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}&=\sum_{\eta\le z < 1}\sum_{n=0}^{\infty}(\bar\eta ^n\bar z)^{-j-1}(\eta^nz)^{s+j-1}\cr
&=\sum_{\eta\le z\le1}\sum_{n=0}^\infty{\bar z}^{-j-1}z^{s-j-1}\epsilon^{n(j+1)}\eta^{n(s+2j)}\cr
&=\left(1-\epsilon^{j+1}\eta^{s+2j}\right)^{-1}\sum_{\eta\le z < 1}{\bar z}^{-j-1}z^{s-j-1}.
\end{array}
$$
Now that the summation has being restricted to an analytic function on $\Re(s) < 1$$z\ge\eta$, giving the desired extension ofterm $\tilde\zeta_\theta$$z^{s-j-1}$ will be bounded, and the sum is guarateed to converge. To see this, substitute back $\mathbb{C}$$z=\lbrace k\theta\rbrace$ and $\bar z=k(\bar\theta-\theta)+\lbrace k\theta\rbrace$. Note that I have arrived atThe summand is of size $\tilde\zeta_\theta$ entire$O(k^{-1-j})$, so thatthe sum converges absolutely for $\zeta_\theta$ only$j\ge1$. For $j=0$ it is $(\bar\theta-\theta)^{-1}\lbrace k\theta\rbrace^{s-1}/k + O(k^{-2})$ for $\lbrace k\theta\rbrace\ge\eta$. As $x^{s-1}$ has poles atfinite variation over the interval $s=1$ and$[\eta,1)$, the sum for $s=2$$j=0$ also converges. There are no poles
Finally, the term $(1-\epsilon^{j+1}\eta^{s+2j})^{-1}$ is meromorphic on $\mathbb{C}$ with zeros on the imaginary axis as suggestedvertical line $-2j+i\mathbb{R}$. So, $\zeta_\theta$ does extend with poles in the questionclaimed set.
