Let $G$ be an abelian group. Is there any probability measure $\mu:\mathcal{P}(G)\to [0,\infty)$ such that for any $A\subseteq G$ and $x \in G$:
$$\mu(A)=\mu(xA)$$
How if $G$ is not abelian? (do you have a counterexample?)
By measure I mean finitely additive measures. (However in Measure Theory measures are infinitely additive).
The fact that abelian groups are amenable (link to MO discussion) is usually proved using the Markov-Kakutani fixed-point theorem. Briefly: The space $\mathfrak{M}(G)$ of finitely additive probability measures on $G$ is a weak*-compact convex subset of the dual space of $\ell^{\infty}(G)$. The group $G$ acts on $\mathfrak{M}(G)$ by continuous affine transformations via $(g\mathfrak{m})(A) = \mathfrak{m}(gA)$ for $g \in G$ and $A \subseteq G$ and if $G$ is abelian, this action on $\mathfrak{M}(G)$ has a fixed-point which, by definition, is an invariant mean.
Finite and solvable groups are amenable groups (see Wikipedia), so amenable groups are not necessarily abelian.
A standard example of a non-amenable group is the free group $F$ on two generators $\{a,b\}$.
Let $W(a)$ be the set of elements of $F$ whose reduced representation as a word in $a^{\pm 1}$ and $b^{\pm 1}$ starts with $a$, similarly for $W(a^{-1}), W(b)$ and $W(b^{-1})$. We can write $F$ as a disjoint union $$F = W(a) \cup aW(a^{-1}) = W(b) \cup bW(b^{-1})$$ and also $$F = \{e\} \cup W(a) \cup W(a^{-1}) \cup W(b) \cup W(b^{-1}).$$ This way of writing $F$ as a disjoint union of two and (up to $\{e\}$) four comparable pieces is sometimes called a paradoxical decomposition because of its relation to the Banach-Tarski paradox.
If $\mathfrak{m}\colon P(F) \to [0,1]$ were an invariant mean, then we would have $\mathfrak{m}(\{e\}) = 0$ (this holds for every infinite group). Since $\mathfrak{m}$ is finitely additive and invariant, we also have $$1 = \mathfrak{m}(F) = \mathfrak{m}(W(a)) + \mathfrak{m}(aW(a^{-1})) = \mathfrak{m}(W(a)) + \mathfrak{m}(W(a^{-1}))$$ and similarly $1 = \mathfrak{m}(W(b)) + \mathfrak{m}(W(b^{-1}))$ from which we derive $$ 1 = \mathfrak{m}(F) = \mathfrak{m}(\{e\}) + [\mathfrak{m}(W(a)) + \mathfrak{m}(W(a^{-1}))]+ [\mathfrak{m}(W(b)) + \mathfrak{m}(W(b^{-1}))] = 0 + 1 + 1 = 2, $$ a contradiction.