Alternatively, we could define $E^G$ to be the set of all functions $f:G\rar E$ such that for all $h\in H$ and all $x\in G$: $\Psi(h)(f(x))=f(xh)$ and $\Psi^G(g)f(x)=f(g^{-1}x)$.
In this case $\Psi^G(g)f\in E^G$ iff $\Psi(h)(Psi^G(g)f(x)) = Psi^G(g)f(xh)$, i.e. $\Psi(h)(f(g^{-1}x)) = f(g^{-1}xh)$, which by definition of $E^G$ comes down to: $f(g^{-1}xh) = f(g^{-1}xh)$.