Convergence for "general" version of Riemann zeta function?

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Let $\{u_k\}_{k\in\mathbb{N}}$ be a non-decreasing sequence of positive reals limiting to infinity.

We define the "general" Riemann zeta sum for $\{u_k\}_k$ by $ \zeta(s) = \sum_{k=1}^{\infty} (u_k)^{-s} $.

We denote by $S \subset \mathbb{C}$ the set of $s \in \mathbb{C}$ for which the sum above converges.

My question is, in general what does $S$ look like? We know for $u_k = k$ that $S$ is a right half-plane $\{\mathrm{Re}s>1\}$. In general can we say $S$ is a right half-plane, to the right of the imaginary axis?

Note: a result like this seems to be used just after equation (1.20) in this paper of Dan Freed. Of course, there are probably more constraints on $\lambda_k = u_k$ in the situation given in the paper than I've put here, but I was wondering how general the result used is. I tried taking a look at the paper which is cited as [Se] after equation (1.20) in Dan Freed's paper, but I couldn't really understand it enough to extract the above statement.

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For $S$ to be a right half-plane, it is sufficient to have $u_k=r_ke^{i\theta}$ for some fixed argument $\theta$, and have moduli $r_k \geq 1$. Then for any $s=x+yi$ we have

$$ \bigg \vert \sum_k u_k^{-s} \bigg \vert = \bigg \vert \sum_k r_k^{-s}e^{-i\theta s} \bigg \vert = \bigg \vert \sum_k r_k^{-x}e^{-yi} e^{-i\theta s} \bigg \vert = \bigg \vert e^{-yi} e^{-i\theta s} \sum_k r_k^{-x} \bigg \vert = \sum_k r_k^{-x}. $$

This shows that the modulus of $\sum_k u_k^{-s}$ is independent of $\text{Im}s$ and decreasing in $x$, which means that the set where the modulus is finite is a right half-plane.

Note that if we drop the requirement that $r_k \geq 1$, we can easily get a left half-plane, e.g. by $u_k=k^{-1}$. I cannot come up with an example of any $u_k$ which don't give a left or right half-plane, so the necessary conditions are likely weaker than mine.