Modulo a (possibly severe) measurability issue, the only groups which admit an invariant submeasure are the amenable groups.

Let $G$ be a non-amenable group and $\mu$ be a $G$-invariant submeasure on $G$. Damien Gaboriau and Russell Lyons showed that every non-amenable group contains a random free subgroup. One way of putting this is to say that there exists a free and measure preserving Borel action of $G$ and $F_2$ on some standard probability space $(X,\alpha)$ and a cocycle
$$c \colon F_2 \times X \to G$$
such that $c_x$ is injective for each $x \in X$ and
$$c(g,x) \cdot x = g\cdot x, \quad \forall g \in F_2,x \in X.$$

Considering now the measure space $Z=G \times X$, we see that the action of $F_2$ on $Z$ given by $g(t,x) = (c(g,x)t,g \cdot x)$ admits a fundamental domain $Y \subset Z$. (The measure of $Y$ is typically not finite. To get a picture, $Y$ plays the role of a set of representatives of cosets in case $F_2$ is an honest subgroup of $G$.)

Let $W$ be a measurable subset of $Z$ and set $$\mu'(W) = \int_X \mu(W_x) d \alpha(x).$$
Moreover, for every element $z$ in the full group of the equivalence relation generated by $G$, we see that $\mu'(zA) = \mu'(A)$. Indeed, the action of each such element on $Z$ is given by a measure preserving shuffle of the $x$-coordinate and an $x$-dependent shift in the $G$-coordinate. Both operations preserve the integral.

Let now $A \subset F_2$ and define
$\mu''(A) := \mu'(A \cdot Y).$ Since $F_2$ lies in the full group of the equivalence relation generated by the action of $G$, we can conclude that $\mu''$ is $F_2$-invariant.

It remains to analyze the implications of the assumption that $\mu$ is exhaustive. Let us assume that there exists an infinite disjoint family $(A_n)$ of subsets of $F_2$ such that $\mu''(A_n) \geq \delta>0$. Then, the functions $x \mapsto \mu(A_{n,x})$ are greater $\delta/2$ on a set of measure at least $\delta/2$. This, with an additional argument, (I did not check every detail) seems to contradict that $\mu$ is exhaustive. Hence, $\mu''$ is exhaustive as well.

This is a contradiction. Hence, every group which admits a $G$-invariant submeasure is amenable.

(The problem with this argument is that $x \mapsto \mu(W_x)$ need not be Borel, so that an additional argument is needed to make sense of the integral in the definition of $\mu'$. Maybe this can be cured, but I do not have time to think about it right now.)