This is probably a trivial observation, but maybe useful to track the difficult part of the problem.

If instead of condition three you require:

(strongly exhaustive) $\lbrace A_n \mid n< \infty\rbrace$ pairwise disjoint implies that $\sum_n \mu(A_n) < \infty.$

then $G$ is amenable. Indeed,

$$\sigma(A) := \sup\left\lbrace \sum \mu(A_i) \mid A = \cup A_i \mbox{ is a finite disjoint union} \right\rbrace$$

defines an invariant mean on $G$. First of all, since $\mu$ is strongly exhaustive, $\sigma(A)$ is a well-defined real number for every $A \subset G$. Secondly, $G$-invariance is obvious from the definition and the $G$-invariance of $\mu$.

Now, if $C = A \cup B$ is a disjoint union, then $\sigma(C) \leq \sigma(A) + \sigma(B)$ follows as before, since each partition of $C$ gives rise to a partition of $A$ and $B$. However, $\sigma(A) + \sigma(B) \leq \sigma(C)$ also holds since a partition of $A$ and a partition of $B$ clearly merge to give a partition of $C$.

All in all, $\sigma(A) + \sigma(B) = \sigma(C)$ whenever $C$ is the disjoint union of $A$ and $B$. This implies that $\sigma$ is a finite $G$-invariant finitely additive measure on $G$, and hence $G$ is amenable.

This is probably a trivial observation, but maybe useful to track the difficult part of the problem.

If instead of condition three you require:

(strongly exhaustive) There exists some constant $C>0$, such that $\lbrace A_n \mid n< \infty\rbrace$ pairwise disjoint implies that $\sum_n \mu(A_n) < C.$

then $G$ is amenable. Indeed,

$$\sigma(A) := \sup\left\lbrace \sum \mu(A_i) \mid A = \cup A_i \mbox{ is a finite disjoint union} \right\rbrace$$

defines an invariant mean on $G$. First of all, since $\mu$ is strongly exhaustive, $\sigma(A)$ is a well-defined real number for every $A \subset G$. Secondly, $G$-invariance is obvious from the definition and the $G$-invariance of $\mu$.

Now, if $C = A \cup B$ is a disjoint union, then $\sigma(C) \leq \sigma(A) + \sigma(B)$ follows as before, since each partition of $C$ gives rise to a partition of $A$ and $B$. However, $\sigma(A) + \sigma(B) \leq \sigma(C)$ also holds since a partition of $A$ and a partition of $B$ clearly merge to give a partition of $C$.

All in all, $\sigma(A) + \sigma(B) = \sigma(C)$ whenever $C$ is the disjoint union of $A$ and $B$. This implies that $\sigma$ is a finite $G$-invariant finitely additive measure on $G$, and hence $G$ is amenable.

This is probably a trivial observation, but maybe useful to track the difficult part of the problem.

If instead of condition three you require:

(strongly exhaustive) $\lbrace A_n \mid n< \infty\rbrace$ pairwise disjoint implies that $\sum_n \mu(A_n) < \infty.$

then $G$ is amenable. Indeed,

$$\sigma(A) := \sup\left\lbrace \sum \mu(A_i) \mid A = \cup A_i \mbox{ is a disjoint union} \right\rbrace$$

defines an invariant mean on $G$. First of all, since $\mu$ is strongly exhaustive, $\sigma(A)$ is a well-defined real number for every $A \subset G$. Secondly, $G$-invariance is obvious from the definition and the $G$-invariance of $\mu$.

Now, if $C = A \cup B$ is a disjoint union, then $\sigma(C) \leq \sigma(A) + \sigma(B)$ follows as before, since each partition of $C$ gives rise to a partition of $A$ and $B$. However, $\sigma(A) + \sigma(B) \leq \sigma(C)$ also holds since a partition of $A$ and a partition of $B$ clearly merge to give a partition of $C$.

All in all, $\sigma(A) + \sigma(B) = \sigma(C)$ whenever $C$ is the disjoint union of $A$ and $B$. This implies that $\sigma$ is a finite $G$-invariant finitely additive measure on $G$, and hence $G$ is amenable.

This is probably a trivial observation, but maybe useful to track the difficult part of the problem.

If instead of condition three you require:

(strongly exhaustive) $\lbrace A_n \mid n< \infty\rbrace$ pairwise disjoint implies that $\sum_n \mu(A_n) < \infty.$

then $G$ is amenable. Indeed,

$$\sigma(A) := \sup\left\lbrace \sum \mu(A_i) \mid A = \cup A_i \mbox{ is a finite disjoint union} \right\rbrace$$

defines an invariant mean on $G$. First of all, since $\mu$ is strongly exhaustive, $\sigma(A)$ is a well-defined real number for every $A \subset G$. Secondly, $G$-invariance is obvious from the definition and the $G$-invariance of $\mu$.

Now, if $C = A \cup B$ is a disjoint union, then $\sigma(C) \leq \sigma(A) + \sigma(B)$ follows as before, since each partition of $C$ gives rise to a partition of $A$ and $B$. However, $\sigma(A) + \sigma(B) \leq \sigma(C)$ also holds since a partition of $A$ and a partition of $B$ clearly merge to give a partition of $C$.

All in all, $\sigma(A) + \sigma(B) = \sigma(C)$ whenever $C$ is the disjoint union of $A$ and $B$. This implies that $\sigma$ is a finite $G$-invariant finitely additive measure on $G$, and hence $G$ is amenable.