Theorem:
Concatenation of two non-regular languages can be regular.
Constructive Proof:
Let $L$ be any non-regular language. Now, we know $L’$ is also non-regular.
Consider $(L \cup \{ \epsilon \})$ ; $(L’ \cup \{ \epsilon \})$, both of which are non-regular.
Now, $(L \cup \{ \epsilon \}).(L’ \cup \{ \epsilon \}) = \Sigma^*$, which is regular.
For concrete example:
Consider $L =$ Set of All Prime Numbers over $\{ 0 \}$ ,
$M =$ Set of All Composite Numbers over $\{ 0 \}$
Both $L,M$ are Non-regular.
Now, $L.M = \{ 0^n , n \geq 6 \}$ which is regular.
My answer here: https://cs.stackexchange.com/a/156523/132177