I hit upon the following identity while reading the book How to Prove: $$(A \cup B) \backslash B \subseteq A$$
Now if I solve this logically I can reduce this like this: $$ \begin{gather*} x \in (A \cup B) \backslash \ B \\ x \in (A \cup B) \land x \in \lnot\ B \\ (x \in A \lor x \in B) \land x \in \lnot\ B \\ x \in \lnot\ B \land (x \in A \lor x \in B) \\ x \in \lnot \ B \land x \in A \end{gather*} $$ The above reduction gives me the intiution that $(A \cup B) \backslash\ B$ is a subset of $A$. But if I think in terms of Venn diagrams, I come up with the following idea: $$(A \cup B) \backslash B = A$$ How to think in these conditions? This fallacy has also been explicity stated in the book.
It is true that if $A\cap B = \varnothing$, then $$(A\cup B) \setminus B = A$$
But we cannot simply assume that $A\cap B$ is empty. That is, $A$ and $B$ need not be disjoint.
Informally, if A and B overlap, then removing all of $B$ from $A\cup B$ removes part of A, too. In this case, $(A\cup B)\setminus B \subsetneq A$.
As for your "logical" argument, you've done fine, assuming by $\lnot B$ you mean the complement of $B$. Otherwise we can argue as follows:
$$\begin{align} x \in (A \cup B) \setminus B &\iff [x \in (A\cup B)] \land (x \notin B)\\ &\iff [x \in A \lor x \in B] \land (x\notin B)\\ &\iff (x\in A \land x \notin B) \lor \underbrace{(x\in B \land x\notin B)}_{\text{false}}\\ &\iff x \in A \land x \notin B\\ &\implies x \in A\end{align}$$ $\therefore \; (A\cup B)\setminus B \subseteq A$.