There is some structure, but I don't think it's reasonable to say there is a classification, without more information.

It's equivalent to ask for a set $S'$ of algebraic integers (each $\geq 1$) such that $\alpha_1,\dots, \alpha_n \in S' \implies \alpha_1,\dots ,\alpha_n, \in \mathbb Q\left(\prod_{i=1}^n \alpha_i\right)$ and then define $S$ to be the set generated under multiplication by $S'$ and $\mathbb N^{ \geq 1}$.

For this construction, removing any element of $S'$ gives another valid $S'$.

If $S'$ is finite, we can also add "generic" elements to $S'$ and preserve the $S'$ property. Given a finite $S'$, choose any number field $K$ (other than $\mathbb Q$), and choose a prime $p$ that splits completely in $K$ and for which all valuations lying over $p$ are trivial on all numbers in $S'$ (possible because there are infinitely many split primes, finitely many numbers in $S'$, and each has finitely many nontrivial valuations). Then take any algebraic integer $\alpha \in \mathcal O_K$ which has positive $\mathfrak p$-adic valuation for some prime $\mathfrak p$ lying over $p$ and zero $\mathfrak p'$-adic valuation for all other primes $\mathfrak p'$ lying over $p$.

If we multiply any positive power of $\alpha$ by any product $\beta$ of integers in $S'$, then $\alpha^n \beta$ still have positive the same positive $\mathfrak p$-adic valuation / zero $\mathfrak p'$-adic valuation property, hence every element of the Galois group that fixes $\alpha^n \beta$ fixes $\mathfrak p$, hence the field $\alpha^n \beta$ generates contains $K$. Because the field $\alpha^n \beta$ generates contains $K$, which contains $\alpha$, it also contains $\alpha^ n \beta/\alpha^n=\beta$, and hence (by the assumption that $S'$ is valid) contains all the elements of $S'$ that multiply to $\beta$.

If we add the condtion $4 \cos (\pi/n)^2 = (\zeta_{2n} + \overline{\zeta}_{2n})^2 \in S'$, this construction won't quite work. It's not too hard to see that the norm of this integer is $p$ if $n$ is $2$ times a power of a prime $p$ and $1$ otherwise, so for each prime $p$, there are some integers with a nontrivial $\mathfrak p$-adic valuation for some $\mathfrak p$ lying over $p$. However, when $n = 2 p^r$, the prime $p$ spits into only one prime in the field $\mathbb Q( \zeta_{2n} + \overline{\zeta}_{2n})$ (because the inertia group of $p$ inside the Galois group of $\mathbb Q(\zeta_{2n})$, which is $(\mathbb Z/4)^\times \times (\mathbb Z/p^r)^\times$, is $(\mathbb Z/p^r)^\times$, and complex conjugation is $-1$, and these generate the whole group). Thus, a slightly modified construction works where we instead choose a prime $p$ such that each element of $S'$ has the same valuation at all primes of the field it generates that lie above $p$.

In general, the key tool to understand such sets will be $p$-adic valuations. Every finitely generated subgroup of a number field is (modulo some roots of unity that can be ignored here) a lattice $\mathbb Z^r$, and subfields define sublattices. So you're looking for a subset of a lattice such that any nonnegative integer combination doesn't intersect a particular sublattice, except if the elements you're taking nonnegative combinations of are already in the sublattice. The obstruction to this happening will be a linear form, vanishing on the sublattice, that takes positive values on all elements lying in that sublattice. Such a form can be written as a linear combination of $p$-adic valuations and logs of absolute values at $\infty$ (corresponding to all the real embeddings of the field, not just one). Using this picture, we can define many such sets $S$ by choosing appropriate linear combinations.

To narrow down what $S$ is, I suspect you'll want to get more arithmetic information, specifically, on the valuations of its elements.