I don't know a reference, but here's the rough proof that $G \in C^1$.

First, from the integral eqaulity
$$
\Phi_t(x+h) - \Phi_t(x) = \int_0^1 D\Phi_t(x+\eta h) \cdot h \,\mathrm{d}\eta
$$
follows the mean-value estimate
$$
\| \Phi_t(x+h) - \Phi_t(x) - D\Phi_t(x) \cdot h \|
\le \sup_{\eta \in [0,1]} \| D\Phi_t(x+\eta h) - D\Phi_t(x) \|\,\|h\|.
$$
This allows us to estimate
$$
\|G(x+h) - G(x) - \int_0^\infty D\Phi_t(x) \cdot h \,\mathrm{d}t\| \\
\le \int_0^\infty \| \sup_{\eta\in[0,1]} D\Phi_t(x+\eta h) - D\Phi_t(x)  \|\,\mathrm{d}t\,\|h\|.
$$
To show that this is $o(h)$, we split up the integral into intervals $[0,T]$ and $[T,\infty)$. For any finite $T$, the first part is small since $t \mapsto D\Phi_t(x)$ is uniformly continuous on $[0,T]$. The second part is exponentially small when $T \to \infty$ due to the exponential stability. Hence $G$ is differentiable and
$$
DG(x) = \int_0^\infty D\Phi_t(x) \,\mathrm{d}t.
$$
Similar estimates can be repeated to prove that $G \in C^1$ and by induction that $G \in C^k$.