Please help.
I have read the prove of the existence of the natural numbers in Analysis I written by Amann and Escher .It uses the claim that the function $\mathcal{V}:\mathbb{N}\longrightarrow\mathbb{N}$, $z\longmapsto z\cup\{z\}$ is injective.Here $\mathbb{N}$ is a set we construct to be the set of natural numbers.I don't know how to prove the claim.
My English is not good.If you don't understand what I said,please tell me.
The answer to the question in the title is yes, assuming the axiom of foundation.
Indeed $A$ is the only set that satisfies "for all $x\in A\cup \{A\}$, $x\in y$ or $x=y$, and $y\in A\cup \{A\}$".
Indeed if $y$ satisfies it, then $y=A$ or $y\in A$. But if $y\in A$, then either $A\in y$ or $A=y$, which is absurd (by the axiom of foundation). Thus $y=A$
However, I believe (I'm not sure about this part) it is consistent to have $A= \{B\}$ and $B=\{A\}$, $A\neq B$ without the axiom of foundation. In such an event, clearly the answer is no.