No you can't use logic operations $\lnot, \land, \lor, etc.$ directly on sets, because sets are not propositions.
For example, when discussing sets themselves, we have $(A \cap \overline{B})\cup(A \cap B)$, and it makes absolutely no sense to translate this simply replace set relations with propositional relations, equating $(A \cap \overline{B})\cup(A \cap B)$ with $(A\land \lnot B)\cup (A \land B)$. The second expression is utterly meaningless
But you can use logic operations/relations on propositions by rewriting a set in terms of what it means for an element to belong to that set, and in that case, we essentially unpack the meaning of set union, set complement, set intersection, etc.
In this case, you have
Let's chase an element in the above set, to unpack what it means; and line by line we can do so entirely with biconditionals:
So $$\begin{align} x \in (A\cap \overline B) \cup (A\cap B) &\iff x \in (A\cap \overline B) \lor x \in (A\cap B)\\ \\
&\iff (x \in A \land x\notin B)\lor (x \in A \land x \in B)\\ \\
&\iff x\in A \land (\underbrace{(x\notin B) \lor (x \in B))}_{\text{tautology} \top}\\ \\
&\iff (x\in A)\land \top\\ \\
&\iff x \in A\end{align}$$
Hence $$x \in (A\cap \overline B) \cup (A\cap B) \iff x \in A$$
Hence, $$\big((A\cap \overline B) \cup (A\cap B)\big) = A$$
Forgetting your misuse of logic relations among sets, you could have saved some time, when you reached this step: $(A ∧ ¬B) ∨ (A ∧ B)$
$(A\cap \overline B) \cup (A\cap B)$ corresponds to $ (x\in A \land x \notin B)\lor (x\in A \land x\in B)$
Then, we can notice that $A$, or rather $x \in A$ appears in both sets on each side of $\cup$ ($\lor$).
So we can use what I sometimes called "reverse distribution", in this case $A \cap (\overline B \cup B)$. Using logic, we have $x \in A \land (\notin B \lor x\in B)$. In terms of sets, we have $A\cap U$, where $U$ is the universal set containing A. In terms of logic, we have that $x \in A \land \top$, where $\top$ means always true for any set: any element is either an element of any given set, or it is not: always true. $A\cap U = A$, and $x\in A \land \top \iff x \in A$
Best Answer
$$\begin{align} \overline{(\overline{A}\cdot\overline{B})+(A\cdot\overline{B})} &=\overline{\overline{A}\cdot\overline{B}}\cdot\overline{A\cdot\overline{B}}\tag{De Morgan's}\\ &=(A+B)\cdot(\overline{A}+B)\tag{De Morgan's}\\ &=A\cdot\overline{A}+A\cdot B+B\cdot\overline{A}+B\cdot B\tag{Expanding}\\ &=0+A\cdot B+B\cdot\overline{A}+B\\ &=B+A\cdot B+B\cdot\overline{A}\tag{Commutative}\\ &=B\cdot(1+A+\overline{A})\tag{Factorise}\\ &=B\cdot1\\ &=B\\ \end{align}$$