A variation of the halting problem
$L_S$ is undecidable if $S = \mathbb{N}_{odd}$:
The problem of whether a given TM accepts $\epsilon$ is undecidable. There is a simple reduction from this problem to the problem of membership in $L_S$ when $S = \mathbb{N}_{odd}$.  Given a TM $M$, we create a new TM $M'$ which on any input simulates $M$ on the same input, but for each step of $M$, it takes one (or any finite odd number of) extra redundant step(s). This basically makes sure that the number of steps taken by $M'$ on any input is always even.
Now, since $T_{M'} \notin S$, hence, $M'$ is in $L_S$ if and only if $M'$ doesn't halts on $\epsilon$, which in turn implies that $M$ doesn't halt on $\epsilon$. Hence, this language is undecidable.
A similar proof can be also given for $\mathbb{N}_{even}$.
[Undecidable $S$ with Oracle Machines]
$L_S$ is decidable for a set $S$ for which $\mathbb{N} \setminus S$ is not finite.
Assuming that the set $S$ need not be decidable itself, and we are going to use it as an oracle and check for decidability using oracle machines.
We construct the set $S$ as follows: for each Turing Machine $M$, let it's binary encoding be $\langle M \rangle$, and let the length of the string $\langle M \rangle$ be $n$. If $M$ doesn't halt on $\epsilon$, then we add the number $10^{n}\langle M \rangle$ to our set $S$. By construction, this set doesn't contain infinite numbers.
Now, the decider will work as follows: On an input $M$, it will check whether the number $10^{n}\langle M \rangle$ is in $S$. It will accept $M$ if the number if found, else we can be sure that the machine $M$ will halt, and hence we simulate $M$ on $\epsilon$ until it halts while keeping the count of the number of steps taken by $M$. We finally accept or reject $M$ on the basis of whether this count is in $S$ or not.
Hence, this language is decidable for a set $S$ for which $\mathbb{N} \setminus S$ is not finite.
[Decidable S]
When the set $S$ is decidable, it will be interesting to know whether the hypothesis that &quot;$\mathbb{N} \setminus S$ is finite&quot; holds. I suspect that the answer would be affirmative.
Let $S' = \mathbb{N} \setminus S$ be infinite set. If $S'$ has &quot;simple&quot; subsequence in it (say, an Arithmetic Progression), then we can prove that $L_S$ would be undecidable by a proof similar to the one in the case of $S = \mathbb{N}_{odd}$. The idea is simply to run a construct a TM $M'$ which simulates $M$, and if $M$ halts, then jump to the next hole in $S$. $M'$ will be in $L_S$ iff $M$ halts.
This idea doesn't seem to work when $S'$ is a difficult set; by which I mean that checking the membership is hard in terms of time complexity.