$\newcommand{\si}{\sigma}
\newcommand{\om}{\omega}
\newcommand{\Om}{\Omega}
\newcommand{\F}{\mathcal{F}}$
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 
\begin{equation}
	\F_A:=\si(\big\{\{C\}\colon C\subseteq A\big\})
	=\si(\big\{\{C\}\colon C\in2^A\big\})=\si(\big\{\{C\}\colon\{C\}\subseteq 2^A\big\}),
\end{equation}
where (for any set $\mathcal E\subseteq2^\Om$) $\si(\mathcal E)$ denotes, as usual, the smallest sigma-algebra $\mathcal G$ over $\Om$ such that $\mathcal G\supseteq\mathcal E$. So, here $\mathcal F_A$ is the smallest sigma-algebra containing/generated by the set of all singleton sets of the form $\{C\}$ with $C\subseteq A$. 
It is easy to see that 
\begin{equation}
	\F_A=\{\Phi\colon\Phi\subseteq2^A\text{ and $\Phi$ is countable}\}
	\cup
\{\Phi\colon\Om\setminus\Phi\subseteq2^A\text{ and $\Om\setminus\Phi$ is countable}\}. 
\end{equation}
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. So, $\zeta$ is not a stopping set. 

----

A counterexample for your second question is as follows. Let $S:=U\times V$, where $U$ and $V$ are any sets of cardinalities $\ge2$ (you may assume that $U$ and $V$ are denumerable and hence $S$ is also denumerable). Let $\Omega:=2^S$, with $\mathcal F:=2^\Om$. For any $\om\in\Om$, let 
\begin{equation}
	\zeta(\om):=p_1(\om)\times V,\quad\eta(\om):=U\times p_2(\om), 
\end{equation}
where 
\begin{equation}
	p_1(\om):=\{u\in U\colon\exists v\in V\ (u,v)\in\om\}\quad\text{and}\quad
	p_2(\om):=\{v\in V\colon\exists u\in U\ (u,v)\in\om\}, 
\end{equation}
the projections of the set $\om$ onto $U$ and $V$. Then 
\begin{equation}
	(\zeta\cap\eta)(\om)=\zeta(\om)\cap\eta(\om)=p_1(\om)\times p_2(\om)
\end{equation}
for all $\om\in\Om$. 
For any $A\subseteq S$, let 
\begin{equation}
	\F_A:=\si(\{\zeta^{-1}(2^C)\colon C\subseteq A\}
	\cup \{\eta^{-1}(2^C)\colon C\subseteq A\}). 
\end{equation}
Then $\mathcal F_A\subseteq\mathcal F_B\subseteq\mathcal F$ for all subsets $A$ and $B$ of $S$ such that $A\subseteq B$. Also, obviously $\zeta^{-1}(2^A)\in\F_A$ and $\eta^{-1}(2^A)\in\F_A$, for any $A\subseteq S$, so that $\zeta$ and $\eta$ are stopping sets. 

Take now any $(x,y)\in S=U\times V$ and let $A$ be the singleton set $\{(x,y)\}$. Then for any $C\subseteq A$ 
\begin{equation}
	\zeta^{-1}(2^C)=\{\om\in\Om\colon p_1(\om)\times V\subseteq C\}=\{\emptyset\}
\end{equation}
and similarly $\eta^{-1}(2^C)=\{\emptyset\}$, whence  
\begin{equation}
	\F_A=\si\big(\big\{\{\emptyset\}\big\}\big)
	=\big\{\emptyset,\{\emptyset\},\Om,\Om\setminus\{\emptyset\}\big\}. 
\end{equation}
On the other hand, the set 
\begin{equation}
	(\zeta\cap\eta)^{-1}(2^A)
	=\{\om\in\Om\colon p_1(\om)\times p_2(\om)\subseteq\{(x,y)\}\}
	=\big\{\emptyset,\{(x,y)\}\big\}  
\end{equation}
is of cardinality $2$ and hence is not in $\F_A$, because none of the four members of $\F_A$ is of cardinality $2$. Thus, $\zeta\cap\eta$ is not a stopping set.