My guess is that the collection $A\equiv\{x|\;\exists y\text{ such that } x\times y = x\}$ is a set, moreover, $A=\{\emptyset\}$
Here is my reasoning: clearly, if we consider $x=\emptyset$, then for every set $y$ we have that $\emptyset\times y=\emptyset$, so $\emptyset\in A$.
However, if $x\not=\emptyset$, it shouldn't be true that $x\times y=x$ for any set $y$, for it should contradict some sort of conclusion of the Axiom of Regularity.
Maybe accepting that $A$ is a set different from $\{\emptyset\}$ would lead us to an infinite sequence $(a_{n})_{n\in\omega}$ of the following form:
Suppose that $x\not=\emptyset$ and $x\in A$. Then there exists $a_0\in x$. But then $a_0\in x\times y$, so $a_0=\langle a_1,b\rangle$ for some $a_1\in x$ (note that $a_0\not=a_1$ because we would be contradicting the Axiom of Regularity, since we cannot have $\{a_0\}\in a_0$ nor $\{a_0,b\}\in a_0$) and $b\in y$ (for if $y=\emptyset$, then $x\times y=\emptyset\not=x$, and $x\not\in A$, contrary to our hypothesis).
Note that from the equality $a_0=\langle a_1,b\rangle$ we can conclude that $\{a_1\}\in a_0$.
But since $a_1\in x$, then $a_1\in x\times y$, and $a_1=\langle a_2, b'\rangle$ for some $a_2\in x$ such that $a_1\not=a_2$ (for analogous reasons to $a_0\not=a_1$), and $b'\in y$ that may or may not be equal to $b$ (since it doesn't really matter).
In conclusion we have that: $\{a_2\}\in a_1\in\bigcup a_0$
This way we can construct an infinite-decreasing sequence (renaming the corresponding sets):
$$\dots\in a'_{n+1}\in a'_n\in\dots\;\in a'_1\in a'_0$$
Following the previous reasoning, which cannot occur in ZF.
Is my proof correct? Any suggestions or insights about it?
Thanks in advance for your time.
Your proof looks right(with only a need to small modifications depends on your definition of 2-tuple).
Note, that although there is no non empty set with that property, for every set $X$ there is a set $Y$ such that $X×X\subseteq Y×Y\subseteq Y$(and you proved that the last inclusion is equality iff $Y=\emptyset\iff X=\emptyset$):
Let $X$ be a set and define $$Y_0=X;\;\;Y_{k+1}=Y_k\cup(Y_k×Y_k);\;\; Y=\bigcup_{k}Y_k$$
First, clearly $X\subseteq Y$, now let $(a,b)\in Y×Y$, then, be definition, $a\in Y_{n_a},\ b\in Y_{n_b}$, WLOG $n_a<n_b$, because we have $Y_m\subseteq Y_{m+1}$ we have $Y_{n_a}\subseteq Y_{n_b}$ so $a\in Y_{n_b}$ as well, so $(a,b)\in Y_{n_b+1}\subseteq Y$ so $(a,b)\in Y$.