Discrete Math Proof With Power-sets The question I am working on is:

Can you conclude that $A=B$ if $A$ and $B$ are two sets with the same power set?

At first, I thought of doing a proof by contraposition; however, that didn't appear (ostensibly, at least) it would prove effective. Although, you of more experience, if you happen to know a way of using that proof method, I would doubtlessly be interested.
So, I then thought of doing a proof by contradiction. Here is my attempt, if you'd be so kind as to look it over and evaluate it, I would appreciate it:
PS I'll have a few questions in bold, sprinkled throughout the proof.
If $A$ and $B$ are two sets with the same power-set, then $A=B$.
Let's assume the opposite is true, that if $A$ and $B$ are two sets with the same power-set, then $A$ and $B$ must be different sets ($A \ne B$).
If $A \ne B$, then either $A$ or $B$ will have at least one element that isn't in the other set. Say $A$ has the extra element, denoted by $e$. (Although you could have $B$ contain the extra element, the argument will work either way, it is independent of the choice).
Let $A= \{a_1,a_2, a_3, ...,e,...\}$, and $B=\{b_1,b_2,b_3,...\}$, such that $B$ doesn't contain $e$ (How would I write that little bit in set notation? Also, do I have to explicitly write that $a_1 =b_1$, $a_2=b_2$,and so on?) 
So, $\mathcal P(A)=\biggr \{ \{a_1\},\{a_2\},\{a_3\},...\{e\},...\{a_1,a_2,a_3,...,e,...\} \biggr \}$
and $\mathcal P(A)=\biggr \{ \{b_1\},\{b_2\},\{b_3\},...\biggr \}$
As you can see, the power-set of $A$ contains at least one element that the power-set of $B$ doesn't have (although, the differ by more than just one). This leads to a contradiction, because we assumed that the power-sets were equal.
 A: Two simple proofs:
(1) Assume $P(A)=P(B)$.  Since $A\in P(A)$, we have $A\in P(B)$, which means $A\subseteq B$.  Similarly, $B\subseteq A$.  Therefore $A=B$.
(2) Every set $X$ is the union of all the members of $P(X)$.  So, if $P(A)=P(B)$, apply to both sides of this equation the operation (often denoted by $\bigcup$)  "union of all the elements of" to get $A=B$.
A: 
Let $A= \{a_1,a_2, a_3, ...,e,...\}$, and $B=\{b_1,b_2,b_3,...\}$, such that $B$ doesn't contain $e$ (How would I write that little bit in set notation? Also, do I have to explicitly write that $a_1 =b_1$, $a_2=b_2$,and so on?)

Rather than worrying about stating that each of $a_i = b_i$, which you would need to do otherwise, why not list:
$A= \{x_1,x_2, x_3, ...,e,...\}$, and $B=\{x_1,x_2,x_3,...\}$, or better yet, $$A= \{x_1,x_2, x_3, ...,e,...\},\;\;B = A\setminus \{e\}.$$
Therefore, if $\mathcal P(A) = \mathcal P(B) \implies A = B$. 
Then you could proceed a bit more easily with your expansion of the power sets of $A$ and $B$.
It might help to conclude that clearly, while $\mathcal{P}(B) \subseteq \mathcal{P}(A)$, $\mathcal{P}(A) \not\subseteq \mathcal{P}(B)$. Hence $\mathcal{P}(A) \neq \mathcal{P}(B)$: contradition!

Indeed, the only element of importance here, is the element $e$, provided $e \in A$, $e\notin B$, hence $\{e\} \in \mathcal{P}(A),\;\land \; \{e\} \notin \mathcal P (B)$...Hence $\mathcal P (A) \not\subset \mathcal P (B),\;$ hence $\mathcal P(A) \neq \mathcal P(B)$, which is a contradiction. 
(Alternatively, it follows by contrapositive from $A \neq B \implies \mathcal P(A) \neq \mathcal P(B)$, if you start from the mere premise that "Suppose $A \neq B$...without loss of generality, we can assume there is some element $e\in A, e \notin B$...)
A: I'll post what I think is a fairly model answer. Notice how I don't need to make reference to any of the other elements in $A$ or $B$, only a single one.
Suppose $P(A)=P(B)$ but that $A\neq B$. Without loss of generality, the statement that $A\neq B$ implies that there exists an element $e\in A$ such that $e\notin B$. It follows that $\{e\}\in P(A)$ and $\{e\}\notin P(B)$. But $P(A)=P(B)$ implies that if $\{e\}\in P(A)$ then $\{e\}\in P(B)$. This is a contradiction and so our assumption that $A\neq B$ is false. Hence $P(A)=P(B)$ implies that $A=B$. QED
