Show that A ∪ (B − A) = A ∪ B by showing each side is a subset of the other side.
= A ∪ (B ∩ A c ) set difference
= A ∪ (A c ∩ B) commutative
= (A ∪ A c ) ∩ (A ∪ B) distributive
= U ∩ (A ∪ B) complement
= A ∪ B identity
Is this the right way to do it?
To elaborate on my comment, in order to prove $A \cup (B - A) = A \cup B$ using the method they requested, we need to pick an element from one side, show that it's in the other side, and then do the same going back from the right to the left side. I encourage you to ponder/prove why $x \in A \Rightarrow x \in B$ means the same thing as $A \subseteq B$.
To demonstrate, to prove $A \cup (B - A) \subseteq A \cup B$, we need to show $x \in A \cup (B - A) \Rightarrow x \in A \cup B$. Recall that to prove a "$p \Rightarrow q$" statement, you suppose $p$ and then show $q$. In that regard, suppose that $x \in A \cup (B - A)$. Then $x \in A$ or $x \in (B - A)$. If $x \in A$, then $x \in A \cup B$ and we are done. If $x \in (B - A)$, then $x \in B \Rightarrow x \in A \cup B$. Either way, we have that $x \in A \cup B$ so that $A \cup (B - A) \subseteq A \cup B$.
For the converse, we need to pick some $x \in A \cup B$ and show that $x \in A \cup (B - A)$. I'll leave this one to you, but once you've done that you'll have shown that $A \cup B \subseteq A \cup (B - A)$. Then since you've shown $A \cup (B - A) \subseteq A \cup B$ and $A \cup (B - A) \subseteq A \cup B$ (i.e., that each side is a subset of the other), you can conclude that $A \cup (B - A) = A \cup B$.
This might seem like a very unintuitive way to prove set equalities at first (indeed your method was much shorter), but it pays off in that this is a go-to method to prove harder set equalities, so it's good to practice it now.