Disjoint union of NP-hard problem and P problem is NP-hard
Given an instance $x$ of $A$, reduce it to an instance $y$ of $A \cup B$ as follows:


Check (in polynomial time) if $x \in B$, if the answer is yes, then $x \not\in A$ since $A \cap B = \emptyset$. Pick any $y \in \Sigma^* \setminus (A \cup B)$ (which is non-empty by hypothesis). 
If $x \not\in B$, then $x \in A \cup B \iff x \in A$. Pick $y=x$.