I claim that *any* infinite set has this property.

Specifically, I'll construct recursively a map $F$ from finite binary strings to finite binary strings with the following properties:

$\sigma\preccurlyeq\tau$ iff $F(\sigma)\preccurlyeq F(\tau)$.

For each $\sigma$, if $f, g$ are two infinite binary strings whose maximal common initial segment is $\sigma$, then $Set(f)\cap Set(g)=\{n<\vert F(\sigma)\vert: F(\sigma)(n)=1\}$.

- Here "$F(h)$" is shorthand for the infinite binary sequence $\bigcup_{\sigma\prec h}f(\sigma)$, and "$Set(h)$" is the set whose characteristic function is $h$.

For each $\sigma$, $\vert F(\sigma)^{-1}(\{1\})\vert\in S$.

Letting $\mathfrak{S}_F=F(f): f\in 2^\mathbb{N}$, we get an uncountable - in fact, size continuum - family of sets whose pairwise intersections always have cardinality lying in $S$.

So now all we need to do is build $F$. But this is a standard finite extension argument, using the fact that $S$ is infinite to say that we always have "enough freedom" to extend the function as desired.

Specifically, the crucial lemma is the following:

We then apply this, over and over, to build $F$: e.g. think of $F(010)$ as being build via a $\sigma$-move off the empty string, and then a $\tau$-move off the resulting string, and then a $\sigma$-move off *that* string.

*Note that this is really just an elaboration of the classical argument: the usual construction of a size-continuum almost disjoint family is just the set of sets of finite binary strings, which in the notation above is $\mathfrak{S}_{id}$. Taking $F=id$ works since we only have to keep getting incomparable extensions. Once we add the "$S$-requirement," though, we need to mix in some additional work, but this requirement can be satisfied by imposing very mild restraints. This type of argument can also be used to construct a perfect set of linearly (or otherwise) independent reals without choice, and - in computability theory - to show that there is a perfect set of Turing-incomparable reals.*

And I think this can be easily tweaked to get the set of cardinalities of intersections to be *exactly* $S$; see my comment below.