I would like to answer this in the negative by providing a counter example. Consider the measurable space $(\mathbb Z, 2^\mathbb Z)$ together with the action by $G := \operatorname{Aut}(\mathbb{Z}) \times \operatorname{Aut}(\mathbb{Z})$ defined by:

$(\gamma,\xi)(n):=
\left\{\begin{array}{ll}
\gamma(n)
& : n \hspace{1.5ex}\mbox{even}
\\
\xi(n)
& : n \hspace{1.5ex}\mbox{odd}
\end{array}\right.$

taking $\gamma$ as a permutation of the even numbers and $\xi$ as a permutation of the odd numbers.

It is not hard to see that the associated type semigroup is isomorphic to $\overline{\mathbb N^2}$ (since we are forced into counting the number of evens and/or odds in a subset of $\mathbb N$). There are precisely four idempotent (types associated to $G$-paradoxical subsets) elements of the type semigroup. They are $\{ \varnothing, (\infty,0),(0,\infty), (\infty,\infty) \}$.

Up to a choice of unit above each of these there aren't many stationary finitely additive measures. They are all parametrized by what they do to the sets $\{1\}$ and $\{0\}$. Any such measure is of the form $\nu(E) = c_1 \times$ (# evens in E) $+ \hspace{1ex} c_2 \times$ (# odds in E) where $c_1,c_2 \in [0,\infty)$.

Consequently, no extension satisfying (a)-(d) exists because (d) would require that for $A = \{1\}, \hspace{1ex} B = \{\mbox{evens}\} \cup \{1\}, \hspace{1ex}$ and $C = \{\mbox{odds}\} \cup \{0\}$ we have constants:

$\nu(A,B) \nu(B,C) = c_2 \times \infty = \infty \neq c_1 = \nu(A,C)$

Comments: As pointed out by the OP (thanks for the catch!) the action I orginally tried to use wasn't a group action. In messing about with this I also ran into some difficulty actually proving that the type semigroup is isomorphic to $\overline{\mathbb{N}^2}$ so I threw the full automorphism group at it so I'd have easy cardinality arguments guaranteeing that fact.