It seems reasonable to guess that any such group $G$ has a map $f:G \to \operatorname{FSym}(\kappa)$ such that $\ker f$ is countable and $s(g) = \operatorname{supp}(f(g))$. This would answer all three questions (yes, no, no).

Some observations that slightly simplify the definition (at least for me): Define a new map $s'$ by $s'(g) = s(g) \setminus s(1)$. Then $s'$ satisfies 1 and 2 and additionally $s'(1) = \emptyset$. Therefore we can assume $s(1) = \emptyset$. Then also $s(g) \mathbin\triangle s(g^{-1}) \subset s(1) = \emptyset$, so $s(g) = s(g^{-1})$. Given $s(g) = s(g^{-1})$, condition 2 simplifies to just $s(gh) \subset s(g) \cup s(h)$. So, in summary, we want a countable-to-one map $s:G \to [\kappa]^{<\omega}$ such that $s(1) = \emptyset$, $s(g^{-1}) = s(g)$, and $s(g h) \subset s(g) \cup s(h)$.

Here $S_\infty$ must be $\mathrm{Sym}(\mathbf N)$ rather than the finitary one, because the finitary one, being countable, is obviously well-behaved in this sense. (Correct me if I misunderstood.)

@LSpice Well he refers to the order $|G|$, so Lie algebra makes sense. It would also make sense to assume $G$ is a finite $p$-group. It hardly makes a difference.

The $k$th term of the lower central series modulo the $(k+1)$th term is generated by the left-normed commutators in $g_1, \dots, g_m$ of weight $k$, of which there are at most $m^{k-1}(m-1)$ for $k > 1$. Now just sum a geometric series and do some estimation, I think. It's obvious it should be $O(m^C)$ anyway.

(Incidentally, wouldn't it be a great feature if the stackexchange system would indicate relevant math.se questions, say with large text overlap, as closely related?)

Crosspost: math.stackexchange.com/questions/4695157/… (it is OK to crosspost after an acceptable period, as you have done, but you should link to the other question)

Since $N$ is normal, the condition in the last line can be simplified to $MM^g N = M^g MN$, i.e., $M$ and $M^g$ commute modulo $N$. So essentially you want to know whether the given condition descends to $G/N$.

Tedious objection: if one of $m,n,r$ is $1$ then the other two have to be equal. Best just to assume $m,n,r>1$.

Since $\alpha$ is irrational, every $p_t(n) = 1/n^t$ for $0 < t \le 1$ can be approximated by some $p_{\{m\alpha\}}$. Therefore it is the same to ask about the span of $\{p_t : 0 < t \le 1\}$.