What is true is that $\tau$ is a map of *modules*; that is,
$$\tau(i^*(x)\cup y) = x\cup \tau(y)$$
for $x\in H^*(G_2)$ and $y\in H^*(G_1)$.

In particular, the kernel of $\tau$ is a sub-$i^*(H^*(G_2))$-module of $H^*(G_1)$.

For an example, consider $G_1=C_p$ (cyclic group) and $G_2=\Sigma_p$ (symmetric group), where $p$ is an odd prime.  The generator $x\in H^2(C_p)$ satisfies $\tau(x)=0$ (since $H^2(\Sigma_p)=0$), but $\tau(x^{p-1})\neq 0$. 

*Added.*  As Neil points out, I'm using cohomology with mod $p$ coefficients here.