Ok I think I've got it. We may bootstrap our way up through conditioning based on functions of the form $\alpha(g') = \alpha(\gamma(g^{-1} \theta)^{-1} g)$ for $\int_G \alpha(g) \eta(\mathrm{d} g) = 1$, and disintegrate the probability measures $\alpha \eta$ through $g' = k_\gamma^\theta$. For each such $\alpha$ and $\theta \in \Theta$, the disintegration theorem of [Leao et al.][1] ensures that $\alpha \eta$ admits a measurable disintegration $(\alpha \eta)_{k_\gamma^{\theta}}(\mathrm{d}g|g')$, i.e., \begin{equation}
    \int_G X(g) \alpha(\gamma(g^{-1} \theta)^{-1} g) \eta(\mathrm{d} g) = \int_G \int_G X(g) (\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g | g') \kappa_\gamma^\theta(\mathrm{d} g'). 
\end{equation}
In particular, for each Borel set $B \in \mathcal{B}(G)$, we may consider $X(g) := 1_B(g) \frac{1}{\alpha(\gamma(g^{-1} \theta)^{-1} g)}$. Consequently,
\begin{equation}
    \eta(B) = \int_G \frac{(\alpha \eta)_{k_\gamma^\theta}(B|g')}{\alpha(g')} \kappa_\gamma^\theta(\mathrm{d} g').
\end{equation}
Therefore any such $\eta_{k_\gamma^\theta}(\mathrm{d} g | g') := \frac{(\alpha \eta)_{k_\gamma^\theta}(\mathrm{d} g|g')}{\alpha(g')}$ can serve as the improper posterior, and this does not depend on specific choice of $\alpha$, up to sets of $\eta$-measure zero. 

Thank you Michael Greinecker for your help! Please let me know if you see any errors.

  [1]: https://www.scielo.cl/scielo.php?pid=S0716-09172004000100002&script=sci_abstract&tlng=pt