Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under atmostcountable unions?

Moti Gitik proved (assuming large cardinals) that all uncountable alephs can have cofinality ω. [All uncountable cardinals can be singular, Israel J. Math. 35 (1980), 61–88] I believe this may be the model you're looking for, but I don't know what happens to nonwellorderable sets in that model. According to the abstract copied below, this model is very close to what you have in mind.
Every wellorderable set $S$ in this model belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions. The proof is by induction on the cardinality of the infinite set $S$. As in the abstract, we may write $S = \bigcup_{i<\omega} S_i$ where $S_i < S$ for each $i$. By the induction hypothesis, each $S_i$ belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions, and thus $S$ belongs to this closure too. Actually, every set $S$ in this model belongs to the closure of $\{\{s\} : s \in S\}$ under iterated countable unions; this is essentially what Gitik's Theorem 6.3 says. 


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 wellorderable 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. 

