If $*$ is a binary operation on a set $X$ then it is costum to define $$ \tag{1}\label{1}A\star B:=\{x\in X:x=a*b\text{ with } (a,b)\in A\times B\} $$ for any $A,B\in\mathcal P(X)$. So it seemed to me (is this correct?) that \eqref{1} defines actually a binary operation on $\mathcal P(X)$ so that I observe that $\star$ is associative or commutative if and only if $*$ is it and moreover if $e$ in $X$ is left or right neutral with respect $*$ then $\{e\}$ is it with respect $\star$: indeed, if $*$ is associative then for any $A$, $B$ and $C$ in $\mathcal P(X)$ the equality $$ A\star(B\star C)=\{x\in X:x=a*y\text{ with }(a,y)\in A\times(B\star C)\}=\\ \{x\in X:x=a*(b*c)\text{ with }(a,(b,c))\in A\times(B\times C)\}=\\ \{x\in X:x=(a*b)*c\text{ with }((a,b),c)\in (A\times B)\times C\}=\\ \{x\in X:x=y*c\text{ with } y\in(A\star B)\times C\}=(A\star B)\star C $$ holds so that $\star$ is associative and analogously if $*$ is commutative then for any $A$ and $B$ in $\mathcal P(X)$ the equality $$ A\star B=\{x\in X:x=a*b\text{ with }(a,b)\in A\times B\}=\\ \{x\in X:x=b*a\text{ with }(b,a)\in B\times A\}=B\star A $$ holds so that $\star$ is commutative; conversely if $\star$ is associative then for any $x_1$, $x_2$ and $x_3$ the equality $$ \{x_1*(x_2*x_3)\}=\{x_1\}\star(\{x_2\}\star\{x_3\})=(\{x_1\}\star\{x_2\})*\{x_3\}=\{(x_1*x_2)*x_3 $$ holds and so even the equality $$ x_1*(x_2*x_3)=(x_1*x_2)*x_3 $$ holds and analogously if $\star$ is commutative then for any $x_1$ and $x_2$ in $X$ the equality $$ \{x_1*x_2\}=\{x_1\}\star\{x_2\}=\{x_2\}\star\{x_1\}=\{x_2*x_1\} $$ holds so that even the equality $$ x_1*x_2=x_2*x_1 $$ holds; finally if $e$ in $X$ is neutral on the left with respect $*$ then for any $A$ in $\mathcal P(X)$ the equality $$ \{e\}\star A=\{x\in X:x=e*a\text{ with }a\in A\}=\{x\in X:x=a\text{ with } a\in A\}=A $$ holds and analogously if it is neutral on the right then the equality $$ A\star\{e\}=\{x\in X:x=a*e\text{ with }a\in A\}=\{x\in X:x=a\text{ with }a\in A\}=A $$ holds. Moreover, I just observed that if $\cdot$ and $+$ are two binary operation in $X$ such that $\cdot$ is left distributive over $+$ and so if $\odot$ and $\oplus$ are the corresponding binary operation on $\mathcal P(X)$ then for any $A$, $B$ and $C$ in $\mathcal P(X)$ the equality $$ A\odot(B\oplus C)=\{x\in X:x=a\cdot y\text{ with }(a,y)\in A\times(B\oplus C)\}=\\ \{x\in X:x=a\cdot(b+c)\text{ with }(a,b,c)\in A\times B\times C\}=\\ \{x\in X:x=(a\cdot b)+(a\cdot c)\text{ with }(a,b,c)\in A\times B\}=\\ \{x\in X:x=y+z\text{ with }(y,z)\in(A\odot B)\times(A\odot C)\}=(A\odot B)\oplus(A\odot C) $$ holds whereas if $\cdot$ is right distributive then by analogous argument the equality holds $$ (A\oplus B)\odot C=(A\odot C)\oplus(B\odot C) $$ holds: so we conclude that $\odot$ is left or right distributive if $\cdot$ is it. Conversely if $\odot$ is left distributive then for any $x_1$, $x_2$ and $x_3$ the equality $$ \{x_1\cdot(x_2+x_3)\}=\{x_1\}\odot\{x_2+x_3\}=\{x_1\}\odot(\{x_2\}\oplus\{x_3\})=\\ \{x_1\}\odot\{x_2\}\oplus\{x_1\}\odot\{x_3\}=\{x_1\cdot x_2\}\oplus\{x_1\cdot x_3\}=\{x_1\cdot x_2+x_1\cdot x_3\} $$ holds so that $\cdot $ is left distributive over $+$ and by analogous argument if $\odot$ is right distributive then even $\cdot$ is it.
So first of all I ask if what I did above is correct and then I ask to prove or disprove that if $E$ in $\mathcal P(X)$ is neutral on the left or on the right with respect $\star$ then $(X,*)$ has not necessarily a left or right neutral element and finally I even ask to prove that if $A$ has left or right inverse with respect $\star$ then the same is not for any $a$ in $A$ with respect $*$. So could someone help me, please?
First, if $X$ is empty, then ${\cal P}(X) = \{\emptyset\}$ and since $\emptyset \star \emptyset = \emptyset$, ${\cal P}(X)$ is a trivial group. Let us leave aside this trivial case and suppose that $X$ is nonempty.
If $E$ is neutral on the left in ${\cal P}(X)$, then in particular, for every $x \in X$, $$ \{x\} = E \star \{x\} = \{ e * x \mid e \in E \} $$ It follows that, for every $e \in E$, $e * x = x$ and thus $e$ is neutral on the left.
Suppose now that ${\cal P}(X)$ has a (two-sided) identity $E$. Then, by the previous argument, one has, for every $e \in E$ and every $s \in X$ $$ e * x = x = x * e $$ It follows that $E$ is a singleton, since, if $e, e' \in E$, then $$ e * e' = e' = e' * e = e $$ Thus $E = \{e\}$ for some $e \in X$ and $e$ is an identity in $X$. Suppose that $A$ has a right inverse $B$. Then $$ \{e\} = AB = \{a * b \mid a \in A, b \in B \} $$ It follows that, for every $a \in A$ and $b \in B$, $a * b = e$. Thus every element of $A$ has a right inverse in $X$.