Give an example of two sets (A and B) for which $(A \cup B) \backslash B \neq A$ - are my answers correct? Basically the only two answers I could come up with are:
1.$A = B$, so combining them and then subtracting $B$ leaves $\emptyset$.
2.$A \subseteq B$, same logic as $1.$, adding them together and then removing $B$ leaves $\emptyset$.
Are there other legit solutions?
 A: We have $(A\cup B)\setminus B\ne A$ if and only if $A\cap B\ne\emptyset$.
By the way your examples are correct only if $A\ne\emptyset$.
A: Since $(A\cup B)\setminus B = (A\cup B)\cap B^c = (A\cap B^c)\cup (B\cap B^c) = A\setminus B$, you have that $$(A\cup B)\setminus B\neq A$$
if and only if $A\setminus B \neq A$. Now, intuitively, $A\setminus B$ is the set of all elements of $A$ which are not in $B$. This will be the same as $A$ if none of the elements of $A$ will be in $B$, do you agree? If so, try to prove this proposition.
A: Your answers are correct if $A$ is not empty, but they are not the only possibilities.
Draw a diagramm of the situation.
Adding $B$ to $A$ and then removing $B$ is the same as just removing all elements of $B$ from $A$.
$(A\cup B)\setminus B = A \setminus B$ 
This will be the same as $A$ iff the intersection is empty, since in that case no elements from $A$ are removed.
Further solutions to your questions are therefore examples where $A\cap B\neq\emptyset$.
For example $A = \{1, 2\}, B = \{2,3\}$.
A: Let's just calculate the answer by simplifying:
\begin{align}
& (A \cup B) \setminus B \;\not=\; A \\
\equiv & \qquad \text{"definition of $\;\ \not\ \;$; set extensionality"} \\
& \lnot \langle \forall x :: x \in (A \cup B) \setminus B \;\equiv\; x \in A \rangle \\
\equiv & \qquad \text{"definition of $\;\setminus\;$; definition of $\;\cup\;$"} \\
& \lnot \langle \forall x :: (x \in A \lor x \in B) \land x \not\in B \;\equiv\; x \in A \rangle \\
\equiv & \qquad \text{"logic: use $\;x \not\in B\;$ on other side of $\;\land\;$"} \\
& \lnot \langle \forall x :: (x \in A \lor \text{false}) \land x \not\in B \;\equiv\; x \in A \rangle \\
\equiv & \qquad \text{"logic: simplify"} \\
& \lnot \langle \forall x :: x \in A \land x \not\in B \;\equiv\; x \in A \rangle \\
\equiv & \qquad \text{"logic: $\;P \land Q \equiv P\;$ and $\;\lnot P \lor Q \;$ are two ways of writing  $\;P \Rightarrow Q\;$"} \\
& \lnot \langle \forall x :: x \not\in A \;\lor\; x \not\in B \rangle \\
\equiv & \qquad \text{"logic: DeMorgan"} \\
& \langle \exists x :: x \in A \;\land\; x \in B \rangle \\
\equiv & \qquad \text{"definition of $\;\cap\;$"} \\
& \langle \exists x :: x \in A \cap B \rangle \\
\equiv & \qquad \text{"basic property of $\;\emptyset\;$"} \\
& A \cap B \not= \emptyset \\
\end{align}
So the answer is: any sets $\;A,B\;$ which have at least one element in common.
