Why does the tensor product of an irreducible representation with the sign representation yield another irreducible representation?

You don't need any character theory to do this. Let $V$ be an irreducible representation of any group $G$ (the group is not necessarily finite and $V$ is not necessarily finite-dimensional) and let $L$ be a $1$-dimensional representation. I claim that $V \otimes L$ is still irreducible. The reason is that tensoring with $L$ is invertible: the natural map $L^{\ast} \otimes L \to 1$ (where $1$ is the trivial representation) is an isomorphism, so

$$(V \otimes L) \otimes L^{\ast} \cong V.$$

Consequently, if $W$ is a proper nonzero submodule of $V \otimes L$, then $W \otimes L^{\ast}$ is a proper nonzero submodule of $V$. More abstractly, tensoring with $L$ is an automorphism of the category of representations of $G$, and automorphisms of categories preserve categorical properties of their objects like irreducibility.

Let $\chi$ be the character of $\rho$ and $\chi'$ be the character of $\rho \otimes \rho_{\Sigma}$. Then observe that $\langle \chi, \chi\rangle=\langle \chi', \chi'\rangle$. But $\langle \chi, \chi\rangle=1$ since $\rho$ is irreducible, so $\langle \chi', \chi'\rangle=1$, and $\chi'$ is irreducible as well.