Is concatenation of regular language with any other language regular?
As you show, if $L = \emptyset$ then $LM$ is regular for any language $M$. Conversely, for languages over $\{0,1\}$ (or any non-unary alphabet), there is a language $M$ such that if $L \neq \emptyset$ then $LM$ is not regular:
$$
M = \{10^p : p \text{ prime}\}.
$$
Indeed, suppose that $w \in L$. Then
$$ w^{-1}(LM \cap w10^*) = M. $$
Since all of the operations preserve regularity, we conclude that $LM$ is not regular.

Over a unary alphabet, the picture is a bit different. Let
$$ M = \{ 0^{n!} : n \geq 0 \}. $$
If $L \neq \emptyset$ is finite then $LM$ is not eventually periodic, and so not regular. If $L$ is infinite and not regular, then $L\{\epsilon\}$ is not regular. If $L$ is infinite and regular, then it is eventually periodic. For every language $M$, $LM$ is also eventually periodic (with the same period), and so regular.