I don't know a reference, but here's the rough proof that $G \in C^1$.
First, from the integral equality
$$
\Phi_t(x+h) - \Phi_t(x) = \int_0^1 D\Phi_t(x+\eta h) \cdot h \,\mathrm{d}\eta
$$
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 $G \in C^k$.
Extra details in reply to Willy Wong's comment:
Since $x^*$ is exponentially stable and $f \in C^1$, it follows that $x^*$ is linearly stable, hence $\|D\Phi_t(x^*)\| \le C e^{-\alpha t}$. Then for $x$ close to $x^*$, $D\Phi_t(x)$ is a small perturbation, hence the exponential estimates are approximately preserved. Since $\Phi_t(x) \to x^*$ and $D\Phi_t(x) = D\Phi_{t-T}(\Phi_T(x)) \cdot D\Phi_T(x)$, we see that when $T$ is large enough, essentially the same estimate holds for any initial $x$. Thus $\|D\Phi_t(x^*)\| \le \tilde{C} e^{-\tilde{\alpha} t}$ with $\tilde{\alpha} \approx \alpha$. This implies that the integral of $D\Phi_t(x^*)$ is well-defined.
Note that the fact that $f \in C^1$ at $x^*$ is also implicitly used to argue that $t \mapsto D\Phi_t(x)$ is uniformly (equi)continuous on any bounded interval.
