= Solution
The displayed identity is false with the printed non-strict inequality. For example, take $A=B=\{y\}$ and let $r=P_y(\widetilde H_y<\infty)\in(0,1)$. The event $L_y\leq\widetilde H_y$ says that the walk returns to $y$ at most once, so its left side is $\mu_y(1-r^2)$, whereas its right side is $e_{\{y\}}(y)=\mu_y(1-r)$.
The standard and evidently intended <last-exit decomposition for a transient random walk> has $L_B<\widetilde H_A$. Decompose that corrected event according to $n=L_B$ and $z=X_n\in B$. The <Strong Markov property> at time $n$ gives
$$
\mu_yP_y(L_B<\widetilde H_A,L_B\geq0)
=\sum_{n\geq0}\sum_{z\in B}
\mu_yP_y(X_n=z,\widetilde H_A>n)P_z(\widetilde H_B=\infty).
$$
Reversibility of the <random walk on a graph> gives the path-reversal identity
$$
\mu_yP_y(X_n=z,\widetilde H_A>n)
=\mu_zP_z(X_n=y,H_A=n).
$$
Since $e_B(z)=\mu_zP_z(\widetilde H_B=\infty)$ is the <equilibrium measure of a finite set>, summing first over $n$ and then over $z$ yields
$$
\boxed{\mu_yP_y(L_B<\widetilde H_A,L_B\geq0)
=\sum_{z\in B}e_B(z)P_z(X_{H_A}=y,H_A<\infty)
=P_{e_B}(X_{H_A}=y,H_A<\infty).}
$$
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