If we assume ZF plus the assertion that $\omega_1$ is regular (which is provable from countable choice), or that indeed there is any uncountable regular cardinal $\delta$, then such a set $S$ exists. (Note that François provided a model having no uncountable regular cardinal, where there is no well-orderable counterexample.)
Let $S$ simply contain elements of Levy rank unbounded in $\delta$. To be specific, you could let $S$ be $\delta$ itself, or $\omega_1$ if this is regular as we usually expect. If a set $X$ has Levy rank $\alpha$, so that $X\in V_{\alpha+1}$, then the union set $\bigcup X$ is also in $V_{\alpha+1}$. In particular, $V_\delta$ is closed under arbitrary unions of its elements. Also, it contains every element of $S$ and also {s} for $s\in S$. Furthermore, since $\delta$ has uncountable cofinality, $V_\delta$ contains as elements all of its countable subsets, since any such subset would have rank bounded below $\delta$. Thus, the clsoure of your set is contained within $V_\delta$, but $S$ is not in $V_\delta$.
A simpler instance of the idea:
The union of any set of ordinals is still an ordinal. Thus, if $\delta$ is an ordinal with uncountable cofinality, then it is already closed under the process of taking countable unions of its elements, but doesn't contain $\delta$ itself as a member.
More specifically, if we take $S=\omega_1$, provided this is regular, then if we start with {s} for all $s\in S$ and close under the process of taking countable unions, we simply get all countable ordinals, and do not generate $\omega_1$ itself this way.