The answer to your first question is negative. E.g., let $\Omega:=2^S=\mathcal P(S)$, with $\mathcal F:=2^\Omega$, and let $\zeta$ be the identity map of $\Omega$, so that $\zeta(A)=A$ for all $A\subseteq S$. For each $A\subseteq S$, let $\mathcal F_A$ be the set of all (at most) countable sets of subsets of $A$ and of the complements to $\Omega$ of all countable sets of subsets of $A$. 
Then $\mathcal F_A$ is the smallest sigma-algebra containing/generated by the set of all singleton sets of the form $\{C\}$ with $C\subseteq A$. Moreover, $\mathcal F_A\subseteq\mathcal F_B\subseteq\mathcal F$ for all subsets $A$ and $B$ of $S$ such that $A\subseteq B$.  

Furthermore, what you denote informally by $\{\zeta=A\}$ is $\zeta^{-1}(\{A\})=\{A\}\in\mathcal F_A$ for all $A\subseteq S$. 

However, what you denote informally by $\{\zeta\subseteq A\}$ is $\zeta^{-1}(2^A)=2^A\notin\mathcal F_A$ for any infinite $A\subsetneq S$, because then neither $2^A$ nor $\Omega\setminus2^A$ is countable.