As defined in C$^∗$-algebras and finite-dimensional approximations by Brown and Ozawa, a discrete countable group $\Gamma$ is bi-exact if its action on $C(\Delta\Gamma):=C(\bar\Gamma)/c_0(\Gamma)$ is amenable, where $ C(\bar\Gamma)=\{f\in\ell^\infty\Gamma\mid f-\rho_t f\in c_0(\Gamma)\ {\rm for\ any}\ s\in\Gamma\}$.
This might turn out to be trivial but I wonder if this is equivalent to requiring the action $\Gamma\curvearrowright\bar\Gamma$ to be amenable. One direction is clear, i.e., if $\bar\Gamma$ is amenable then so is $\Delta\Gamma$. For the converse, as $\bar\Gamma=\Gamma\sqcup \Delta\Gamma$, one may define $\mu:\Gamma\ni t\to\delta_t\in {\rm Prob}(\Gamma)$ and then "paste" it on the sequence of continuous maps $\mu_n:\Delta\Gamma\to{\rm Prob}(\Gamma)$ coming from the amenable action $\Gamma\curvearrowright\Delta\Gamma$. However, it's unclear to me whether the resulting maps $\mu_n: \bar\Gamma\to{\rm Prob}(\Gamma)$ is continuous or not.
