Yes – if the players are allowed to mix with finitely additive probability measures ( “charges’’ on the power set of F₂ ) the game always has a saddle point, hence a Nash equilibrium, no matter what subset
W ⊂ F₂ is chosen. In particular the phenomenon has nothing to do with amenability; it holds for every group and for every infinite two–person zero–sum matrix game once finitely additive mixed strategies are admitted.

Why it is always solvable

1. Strategy spaces.
   Let M := ba₁(F₂) be the set of all finitely additive probability measures on F₂.
   Identify M with the set of positive norm–1 functionals on the Banach space
   ℓ^∞(F₂). Equipped with the weak*‐topology coming from ℓ^∞(F₂)*, M is compact (Banach–Alaoglu) and convex.

2. Pay–off function.
   For x,y ∈ F₂ put
   k_W(x,y) = χ_W(xy) ∈ {0,1}.
   For μ,ν ∈ M define

   f_W(μ,ν) = ∫_F₂ ν(dy) ∫_F₂ χ_W(xy) μ(dx)      (1)
   

(The inner integral is just the value of μ on the bounded function
x ↦ χ_W(xy); the outer one is the value of ν on the bounded function
y ↦ μ(χ_W(∙ y)).)
Because the integrand is bounded by 1, (1) makes sense for every pair μ, ν, and

    0 ≤ f_W(μ,ν) ≤ 1 .

Linear algebraically, (μ,ν) ↦ f_W(μ,ν) is affine (indeed bilinear) in each argument and is weak*–continuous in each variable separately.

3. Minimax theorem.
   Let X = Y = M and let g(μ,ν)=f_W(μ,ν).
   • X and Y are compact and convex in the locally convex weak* topology.
   • g is linear (hence convex in ν and concave in μ) and jointly bounded.
   Hence Fan’s / Sion’s minimax theorem (or Glicksberg’s generalisation of von Neumann’s theorem) applies:

   max_{μ∈M} min_{ν∈M} g(μ,ν)  =  min_{ν∈M} max_{μ∈M} g(μ,ν) = v.
   

Moreover there exists at least one pair (μ̄, ν̄) ∈ M×M (a saddle point) such that for every μ,ν ∈ M

    g(μ,ν̄) ≤ g(μ̄,ν̄) ≤ g(μ̄,ν).                      (2)

4. Translating (2) back to the integrals gives exactly

    ∫∫ χ_W(xy) dμ(x) dν̄(y) ≤ ∫∫ χ_W(xy) dμ̄(x) dν̄(y)
        ≤ ∫∫ χ_W(xy) dμ̄(x) dν(y)

for all finitely additive μ, ν, i.e. the inequalities you asked for.
Therefore (μ̄, ν̄) is a Nash equilibrium of the multiplication game with pay-off set W.

Consequences and remarks

• The construction requires no invariance properties of μ̄ or ν̄, so amenability plays no rôle. (For amenable groups one can pick right– or bi–invariant means and get one pair that works simultaneously for every W, but that is a different question.)

• The same argument gives a value and an equilibrium for ANY two–player zero–sum game with an arbitrary bounded matrix (possibly uncountable) once finitely additive mixtures are permitted. Classical counter-examples of Shapley and others show that such games need not have a value if one insists on countably additive strategies; finitely additive strategies restore the minimax property.

• Thus the “multiplication game’’ does not characterise amenability: it is solvable on every group, including free groups and other non-amenable ones.