Does $\Sigma^* \cdot a^nb^n=\Sigma^*$
Yes if $\Sigma \supseteq \{a, b\}$. To show the equality, let's show one is a subset of the other, and vice versa.


$\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \subseteq \Sigma^\ast$ holds, because for any word $w \in \Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\}$, we know $w = u a^n b^n$ for $u \in \Sigma^\ast$, therefore $w \in \Sigma^\ast$ since $\Sigma \supseteq \{a, b\}$.
$\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast$ holds, because

$$
  \Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast \cdot \{ a^0 b^0 \} = \Sigma^\ast \cdot \{ \varepsilon \} = \Sigma^\ast
$$

where $\varepsilon$ is the empty word.