I think it is sufficient to assume that $G$ is a connected Lie group. It is not necessary to assume that the exponential map is surjective (intuitively because one can follow broken one-parameter groups).

In fact, the set $H$ consisting of all $g\in G$ such that $\Phi_g^*\omega=\omega$ is a closed subgroup of $G$. A closed subgroup of a Lie group is a Lie group with the induced topology. What is the Lie algebra of $H$? The assumption that $L_{\phi_\chi}(\omega)=0$ for all $\chi\in\mathfrak g$ implies that the Lie algebra of $H$ equals the Lie algebra of $G$. In case $G$ is connected, this implies that $H=G$.

I think it is sufficient to assume that $G$ is a connected Lie group. It is not necessary to assume that the exponential map is surjective (intuitively because one can follow broken one-parameter groups).

In fact, the set $H$ consisting of all $g\in G$ such that $\Phi_g^*\omega=\omega$ is a closed subgroup of $G$. A closed subgroup of a Lie group is a Lie group with the induced topology. What is the Lie algebra of $H$? The assumption that implies that the Lie algebra of $H$ equals the Lie algebra of $G$. In case $G$ is connected, this implies that $H=G$.

Edit: Another way to think, perhaps more elementary. $H$ is an (abstract) group. The assumption that $L_{\phi_\chi}(\omega)=0$ for all $\chi\in\mathfrak g$ implies that $H$ contains a neighborhood of the identity, and $G$ (if connected) is generated by any neighborhood of the identity.

To see that $H$ contains a neighborhood of the identity, one uses that the exponential map is onto a neighborhood of the identity and computes that (denote $\varphi_t=\Phi_{e^{t\chi}}$):

$$\frac{d}{dt}\Big|_{t=s}\varphi_t^*\omega=\frac{d}{dt}\Big|_{t=s}(\varphi_{t-s}\varphi_s)^*\omega=\frac{d}{du}\Big|_{u=0}(\varphi_u\varphi_s)^*\omega=\varphi_s^*\frac{d}{dt}\Big|_{u=0}\varphi^*_u\omega=0$$

for all $s\in\mathbb R$, so $\varphi_t^*\omega=\omega$ for all $t\in\mathbb R$.