Showing Context Free Grammars that are regular if and only if L(G) = Σ∗.
This proof-snippet is what you are probably looking for.

We follow your definition. $N$ is a fixed non-regular context-free language over $\Sigma$, $\#$ is a new symbol. For a given grammar $G$ let $L = N\#\Sigma^* \cup \Sigma^* \# L(G)$. Assume $L(G) = \Sigma^*$, then $L = \Sigma^*\#\Sigma^*$ which is a regular language. On the other hand, if $w\notin L(G)$, then $L\cap \Sigma^*\#w = N\#w $, which is not regular (as regular languages are closed under quotient, and $N$ is not regular). Thus also $L$ itself is not regular (as regular languages are closed under intersection).

Conclusion: $L$ is context-free, and $L$ is regular iff $L(G) = \Sigma^*$.