Keeping $L_2$ as $\Sigma^*,$ what ever may be $L_1,$, we get a Regular language.
So, statement I is wrong.
If regular languages are closed under infinite union, then $L=\{a^n.b^n | n>0 \}$ must be regular as it is equal to $\{ab\} \cup \{aabb\} \cup \{aaabbb\} \cup \ldots$
So, statement II is wrong.
Option D is correct.