You need the extra requirement that the group element $g$ is non-trivial. Then the Plancherel Theorem implies that the right hand side equals the trace of the left translation $L_g$ on $\ell^2(G)$. Then you compute this trace as $$ \mathrm{tr}(L_g)=\sum_{y\in G}\langle L_g\delta_y,\delta_y\rangle=\sum_{y\in G}\langle \delta_{gy},\delta_y\rangle=0. $$

The assertion holds with deg$(\rho)^2$ replaced with deg$(\rho)$. You also need the extra requirement that the group element $g$ is non-trivial. Then the Plancherel Theorem implies that the right hand side equals the trace of the left translation $L_g$ on $\ell^2(G)$. Then you compute this trace as $$ \mathrm{tr}(L_g)=\sum_{y\in G}\langle L_g\delta_y,\delta_y\rangle=\sum_{y\in G}\langle \delta_{gy},\delta_y\rangle=0. $$