It is possible to have $S$ and a matrix $B$ for which $B$ does not commute with anything in $S$, but $B$ does commute with the limit of a sequence of finite products of things in $S$. In fact, we'll find a sequence with limit $A$ for which $A$ doesn't commute with any finite products of things in $S$ (but, of course, $A$ commutes with itself).
Let $S=\{e^{2\pi i\sqrt{2}}, e^{2\pi j \sqrt{2}}\}\subseteq Sp(1) = $ unit quaternions. I claim that the closure of the group generated by $S$, denoted $\overline{S}$ is all of $Sp(1)$.
The idea is that just by looking at $e^{2\pi i\sqrt{2}}$, the closure of the smallest subgroup containing this is an $S^1 = \{e^{i\theta}\}\subseteq Sp(1)$ and likewise the other element generates another $S^1 = \{e^{j\theta}\}\subseteq Sp(1)$.
Hence, $\overline{S}$ must contain these two circles. Note that since $\overline{S}$ is a closed subgroup of $Sp(1)$, it's an embedded Lie subgroup, so we can talk about its Lie algebra. On the Lie algebra level, this implies the Lie algebra of $\overline{S}$ contains $i$ and $j$. Since it's a Lie subalgebra, it contains $[i,j] = 2k$. Since its a subspace, it must contain all combinations of $i,j,k$, i.e., it's all of $\mathfrak{sp}(1)$.
This implies $\overline{S} = Sp(1)$.
Now, it's easy to find things which don't commute with any element in $S$. For example, $k$ doesn't since $k e^{2\pi i \sqrt{2}} k^{-1} = e^{-2\pi i \sqrt{2}}$ and likewise for $e^{2\pi j \sqrt{2}}$. On the other hand, of course, $k$ commutes with itself.
If you're more stringent and want to find an element of $Sp(1)$ which doesn't commute with any finite combination of things in $S$, you can still do it (I think).
The idea is that the group generated by $S$ is countable. Each element of $Sp(1)$, with the exception of $\pm 1$ lie in a unique maximal torus. I think one can prove (but I admit I haven't gone through the details) that no finite combination of things in $S$ produces $\pm 1$. (If some finite combination of things in $S$ does make $\pm 1$, then I'd bet that changing one of the $\sqrt{2}$s to something algebraically independent from $\mathbb{Q}(\sqrt{2})$ would fix it.) Believing this, now argue as follows.
For each finite combination of things in $S$, the set of all matrices which commutes with this is an $S^1\subseteq Sp(1)$. Doing this for all finite combinations of things in $S$ gives countably many $S^1\subseteq Sp(1)$. Since there are uncountably many $S^1\subseteq Sp(1)$, at least one $S^1\subseteq Sp(1)$ has not been used. Pick any element in this $S^1$ which is not $\pm 1$. This element commutes with itself, but not with any finite combination of things in $S$.