Yes:  Let $\alpha$ be a cohomology class in $Y$.  For simplicity, let's assume that $PD_{Y}\alpha$ is represented by an embedded cycle in $Y$.  Then $PD_{X}\pi^{*}\alpha=\pi^{-1}PD_{Y}(\alpha)$.  Therefore, $\pi\circ PD_{X}\pi^{*}\alpha = \pi (\pi^{-1}PD_{Y}\alpha)=|G|PD_{Y}\alpha$ as the map $\pi: \pi^{-1} PD_{Y}\alpha\rightarrow PD_{Y}\alpha$ is a $|G|$-to-1 covering.  Now apply $PD_{Y}^{-1}$ to the previous equation to obtain $\pi_{!}\circ \pi^{*}\alpha=|G|\alpha$.