I have two formulas. The first one with free variables $\{S\}$ is:
$$\forall x(x \in S \Rightarrow P(x))$$
The second one with free variables $\{x,S\}$ is:
$$x \in S \Rightarrow P(x)$$
I guess we do not have in first order logic:
$$\vDash \forall x(x \in S \Rightarrow P(x)) \Leftrightarrow (x \in S \Rightarrow P(x))$$
Here is an attempt of mine. I use a model where $P=\{<0>\}$. And then I find that for these assignments I get different truth values:
$$\begin{array}{ccc} x & S & \forall x(x \in S \Rightarrow P(x)) \\ \hline 0 & \{1\} & F \\ 0 & \{0,1\} & F \end{array}$$
$$\begin{array}{ccc} x & S & x \in S \Rightarrow P(x) \\ \hline 0 & \{1\} & T \\ 0 & \{0,1\} & T \end{array}$$
Is this a way to show semantical inequality?
They are not semantically equivalent in that -- for any given interpretation of $S$ and $P$ -- the first has only one truth value, whereas the truth value of the second depends on which value you give $x$.
For example, in your model with $S=\{0,1\}$ and $P$ being true for $0$ only, then your second formula is true sometimes, namely when $x=0$ (and in fact whenever $x\ne 1$) -- but with the same interpretation the first formula is simply false, period.