I don't understand the author's argument in the second line of the proof. In particular, I don't see exactly how such a bijection does exist.
I see that $ab \in L_b$ implies $H_b = H_{ab}$, and the bijection goes from $H_b$, but don't see how it goes onto $L_a \cap R_b$.
By Green's Lemma, $\rho_c$ induces an $\mathcal R$-class preserving bijection from $L_b$ to $L_{bc}$. In particular $bc \in R_b$. Moreover since $L_{ab} = L_b$ and $\rho_c(ab) = abc = a\,$, one gets $L_{bc} = \rho_c(L_b) = \rho_c(L_{ab}) = L_{abc} = L_a\,$. Thus $bc \in L_a$ and finally $bc \in L_a \cap R_b$.