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 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.