There are lots of different questions going on here, but here are a few observations.

First of all, if you change $V$ by any set with (upper) density zero, then $c_V(t)$ doesn't change for any value of $t$. So for any $T_V$ there are uncountably many possible choices of $V$, and so almost all of them will be transcendental functions. (For an explicit example where $T_V = [0,1)$, let $V_S$ consist of the set of all integers minus any subset $S$ consisting of powers of $2$. The set of possible $S$ is uncountable.)

Second, it's also relatively easy to construct $V$ such that $T_V$ is missing infinitely many rational numbers. Let's start out with an example which is just missing $1/3$ and $2/3$. The idea is to take $V$ to consist of a string of numbers which are $1$ mod $3$, then a much longer string of numbers which is $2$ mod $3$, then a much much longer string which is $1$ mod $3$ and so on. That is,

$$\varsigma(x) = \frac{x}{1-x^3} + x^{3a_1} \frac{x^2 - x}{1-x^3} + x^{3 a_2} \frac{x - x^2}{1 - x^3} + x^{3 a_3} \frac{x^2 - x}{1-x^3} + \ldots $$

Note that

$$\varsigma(x) = \frac{x}{1-x^3} \left(1 + (x-1) x^{3a_1} - (x-1) x^{3 a_2} + (x-1) x^{3 a_3} + \ldots \right)$$

Here we imagine that the $a_n$ are growing really really fast, (say $x = n!^3$ or something).
Let $t = 1/3$ and let $\omega = e^{2 \pi i t}$.
Then $c_V(1/3)$ exists if and only if

$$\lim_{x \rightarrow 1^{-}} \left(1 + (\omega x-1) x^{3a_1} - (\omega x-1) x^{3 a_2} + (\omega x-1) x^{3 a_3} + \ldots \right)$$

exists. The claim is that the value of this sum will vary between $1$ and $\omega$ as $x$ approaches $1$. The reason is that when $a_n$ increases sufficiently fast, one can choose $x$ so that the first $n$ values of $x^{3 a_n}$ are very close to one and the others decrease super-exponentially. For example,
choose a value of $x$ such that

$$1 - \frac{1}{n^2 a_n} < x < 1 - \frac{n}{a_{n+1}} < 1 - \frac{n^{k}}{a_{n+k}}.$$

Then for $m \le n$,

$$x^{3 a_m} \ge x^{3 a_n} \ge \left(1 - \frac{1}{n^2 a_n}\right)^{3 a_n}
\sim 1 + O\left(\frac{1}{n^2}\right),$$

but for $m = n+k > n$,

$$x^{3 a_m} \le \left(1 - \frac{3 n^{k}}{a_{n+k}}\right)^{3 a_{n+k}}
\sim e^{-3 n^k}$$

which are thus negligible. Thus the value of the sum is approximately

$$1 + (\omega - 1) - (\omega -1) + (\omega -1) - \ldots $$

and so will vary between $1$ and $\omega$ and not converge.

On the other hand, it's not so hard to see that there will be no convergence issues for all *other* rational numbers except for $t = 1/3$ and $t = 2/3$. The point is that the $1-x$ term kills of any power series with terms that are so spread out.

But now consider:

$$\varsigma_W(x):=\varsigma_V(x) + \varsigma_V(x^3) + \varsigma_V(x^9) + \ldots $$

Since $V$ consisted only of terms $1$ and $2$ modulo $3$, there exists a corresponding set $W$. But now the analysis above shows that $T_W$ avoids the set $a/3^n$ for every $(a,3) = 1$ and $n \ge 1$. The point is that these values are screwed up exactly by $\sigma_V(x^{3^{n-1}})$. So $T_W$ is missing infinitely many rational numbers.

Some preliminary analysis suggests one can construct functions missing all rational numbers except zero but the analysis could be a bit painful. (The idea is similar to the above, but to slowly increase the modulus one is considering from $2!$ to $3!$ to $4!$ etc.)

Other observations:

Certainly a $V$ chosen *randomly* will, almost always (in the appropriate sense) have $S \subset T_V$ for any finite set $S$. I might even guess that for a random $V$ the set $T_V$ contains all rational numbers, did you try to prove that?

adoreattention, so, the more the better. xD This is all for my doctoral dissertation. My challenge is two-fold: (1) no one at my university is doing anything close to analytic number theory; (2) I"m doing mathematics of the problem-solving kind, rather than the theory-building, so, finding the specific information applicable to my circumstances is often arduous. Asking about things online has been a life-saver. $\endgroup$p-adic analysis, of all things!) I've found a generalization of my initial result that appears to work for all functions in the class I describe in italics at the end of my question. It's nerve-wracking, being stuck in the dark like this as to the significance my work. Hopefully, though, adding a bounty to this question will get me an answer and put me out of my misery. ;) $\endgroup$2more comments