Here's a heuristic that suggests that the density might matter. It's not always right, as Jeremy has pointed out.

**Fundamental assumption:** Let $\mathrm{den}(P)$ denote the denominator of $P$, and assume that $\mathrm{den}(mP)$ is essentially a random integer of its size, subject to the condition that $\mathrm{den}(Q)\mid\mathrm{den}(kQ)$ for any point $Q$ and integer $k$. This can be broken by choosing the set $S$ to depend on the given elliptic curve (c.f. Jeremy's answer), but if the set $S$ has no relation to the curve, then this might be plausible.

I also want to assume, given some point $P$, that $\log \mathrm{den}(mP) \asymp m^2$. This is reasonable by a consideration of heights. I'm also going to ignore the divisibility condition, since this also shows that its contribution to $\log(\mathrm{den}(mP))$ should be smaller by a factor of a constant.

For simplicity, let's assume that $E(\mathbb{Q})$ is infinite cyclic, and let $Q$ denote a generator. For a number of size $N$, the probability that $N$ is composed only of primes in $S$ is $1/(\log N)^{1-\alpha}$, where $\alpha$ is the density of $S$. If we look at primes up to some point $X$, then, believing that $\mathrm{den}(pQ)$ is random of its size, the expected number of $m\leq X$ for which $\mathrm{den}(mQ)$ is $S$-integral should be
$$
\sum_{m\leq X} \frac{1}{(\log \mathrm{den}(mQ))^{1-\alpha}} \asymp \sum_{m\leq X} \frac{1}{m^{2(1-\alpha)}}.
$$
Notice that if $\alpha>1/2$, this sum diverges as $X\to\infty$, indicating we should expect there to be infinitely many $S$-integral points. On the other hand, if $\alpha<1/2$, the sum converges, and we should expect only finitely many $S$-integral $pQ$'s. In the unlucky situation that $\alpha=1/2$, though the sum diverges, what I've said is fuzzy enough that I wouldn't be comfortable making a guess here (and perhaps, both behaviors can occur, even if the assumption holds).

order problemand has been studied recently. A first reference could be the papers of Antonella Perrucca, who quotes earlier results of Khare, Prasad, Pink, Silverman, etc. $\endgroup$