Question: Can it be that there exists some infinite set $S$ and some binary operation $b: S \times S \to S$ such that cardinality of subsets of $S$ which are closed under $b$ is greater than the cardinality of $S$?

*Edit*: Since edit was requested, I would like to mention that I wanted $b$ to include both of the variables in its definition and that it is almost-everywhere non-constant (in other words, it is possibly constant only on finite number of pairs) so that binary operations such as $b(s\_1,s\_2)=s\_1$ and $b(s\_1,s\_2)=s\_\*$ are not allowed.

Answer: 
In fact the statement is true in both directions.

1. Let $y\_n\rightharpoonup y$ in $X$ i.e. $\forall\varphi\in X^\ast$ $\varphi(y\_n)\to\varphi(y)$. Supposing $\exists\varphi\_0\in Y^\ast$ such that $\varphi\_0(y\_n)\not\to\varphi\_0(y)$. By Hahn-Banach there exists $\psi\in X^\ast$ such that $\psi(Y)=\varphi\_0$. Thus $\psi(y\_n)=\varphi\_0(y\_n)\not\to\varphi\_0(y)=\psi(y)$. A contradiction.
2. Let $y\_n\rightharpoonup y$ in $Y$ i.e. $\forall\varphi\in Y^\ast$ $\varphi(y\_n)\to\varphi(y)$. Supposing $\exists\varphi\_0\in X^\ast$ such that $\varphi\_0(y\_n)\not\to\varphi\_0(y)$. Then there exists $\psi=\varphi\_0(Y)\in Y^\ast$ and $\psi(y\_n)=\varphi\_0(y\_n)\not\to\varphi\_0(y)=\psi(y)$. A contradiction.
Yes, this is the case. 

Key Lemma: Let $X$ be a normed space and $Y \subseteq X$ a linear subspace, then $X^\* \subseteq Y^\*$.

proof: Let $f \in X^\*$, then $f: X \rightarrow \mathbb{C}$ linear and $\Vert f \Vert\_{X^\*} := \sup\_{x \in X} \frac{\vert f(x) \vert}{\Vert x \Vert} < \infty$. Then since $Y \subseteq X$, the restriction of $f$ to $Y$ is also linear and we have $\Vert f \Vert\_{Y^\*} := \sup\_{x \in Y} \frac{\vert f(x) \vert}{\Vert x \Vert} \leq \sup\_{x \in X} \frac{\vert f(x) \vert}{\Vert x \Vert} < \infty$ i.e. $f \in Y^\*$. 

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Now assume $(y\_n)$ converges weakly on $Y$, then we have $\forall f \in Y^\*: f(y\_n) \rightarrow f(y)$ and thus by the lemma: $\forall f \in X^\*: f(y\_n) \rightarrow f(y)$ i.e. weak convergence in $X$.