Let $\mathfrak{A}$ be a Boolean algebra and $E$ be an element in $\mathfrak{A}$. The set of all subelements of $E$ forms a Boolean algebra, denoted by $\mathfrak{A}_E$. Suppose that $I$ be the principal ideal (in $\mathfrak{A}$) generated by $E^c$. That is:
$$I=\{A\in\mathfrak{A}: A\cap E=\emptyset\}$$
Then, the algebra $\mathfrak{A}_E$ is isomorphic to $\mathfrak{A}/I$.
Q1/ Are elements in $\mathfrak{A}_E$ of this form $$A\in\mathfrak{A}\Longleftrightarrow A\cap E\in \mathfrak{A}_E$$
and if $A,B\in\mathfrak{A}$, so $(A\cap E)\cap(B\cap E)=(A\cap B)\cap E=A\cap_E B\in\mathfrak{A}_E$. Am I correct ?
Q2/ How to show that the above two algebras are ISO.
I must admit that I do not really understand your first question. However, if you meant to ask if $\mathfrak{A}_E$ is equal to the set $\{ A \cap E \:|\: A \in \mathfrak{A} \} =: \mathfrak{A}_E'$, then you are right, because any element $B \in \mathfrak{A}_E$ satisfies $B = B \cap E \in \mathfrak{A}_E'$ and conversely any element $B$ of $\mathfrak{A}_E'$ satisfies $B \cap E = B$ which implies $B \in \mathfrak{A}_E$.
Your computation afterwards is correct as far as I can tell, at least as long as the last equality is the definition of $A \cap_E B$.
For your second question you can look at the map $\mathfrak{A} \rightarrow \mathfrak{A}_E$ mapping an element $A \in \mathfrak{A}$ to $A \cap E \in \mathfrak{A}_E$. By the above, this is a well-defined and surjective map. Now, your (second) definition of $I$ shows that $I$ is precisely the kernel of this map and you are done by the homomorphism theorem.
Notice that this situation is a special case of the Peirce decomposition of an algebra by a central idempotent.