Consider two sets $A$ and $e$ such that their symmetric difference $A\Delta e= A$. That implies $A\cap e= \emptyset$, which I tried to prove like, $A\Delta e= A\Rightarrow A\subseteq A\Delta e\Rightarrow (x\in A\Rightarrow x\in A\Delta e)$. Now $x\in A\Delta e$ means if $x\in A$, which is the case here, $x\notin e\Rightarrow A\cap e= \emptyset$. Correct me if I am wrong anywhere in the proof and also please provide any other methods to do the above.
Next the author claims that $A\Delta e= A$ implies $e\subseteq A$, which I can't prove. I tried the problem with help of venn diagrams, if I suppose that $e\subseteq A$, then $A\Delta e$ which is given by $(A\cup e)-(A\cap e)$, with $A\cap e= e$ and also $A\cup e= A$ thus we have $A-e \neq A$, since $A\nsubseteq A-e$ . In other words, the common portion of $A$ and $e$ which is essentially $e$ is pulled off.
I will write an answer so that this question gets marked as answered.
As I said in the comment, we have $A\cap e=\emptyset$ and $A\cap e=e$, which imply $e=\emptyset$. The claim $e\subseteq A$ follows.
Have a good day. :)