If $L$ is undedidable and $A$ is finite, then $L \cup A$ is undecidable.
To see this notice that, given $w \in \Sigma^*$:
$$w \in L \iff (w \in L \cap A) \; \vee \; (w \in L \cup A \;\wedge\; w\not\in A).$$
Since $A$ if finite both $A$ and $L \cap A$ (which is also finite) are decidable. That is, we can test whether $w \in L \cap A$ and whether $w \not\in A$.
If $L \cup A$ were decidable the above formula would then imply that $L$ must also be decidable, a contradiction.
If $L$ is undedidable and $B$ is decidable, then $L \cup B$ might or might not be undecidable, depending on $B$.
An example in which $L \cup B$ is undecidable is the one discussed before: let $B$ be any finite language, e.g., $B=\emptyset$.
An example in which $L \cup B$ is decidable is $B=\Sigma^*$ since $L \cup B = \Sigma^*$.