Buffer Systems and Ion Reaction
Yes the equilibrium certainly exists. In fact, there exists a third equilibrium, the self ionisation of water. Considering any of the two equilibria gives the concentrations of all the species involved, there is no need to write down the third equilibrium as the third equation will be a combination of the first two. The problem arises only when you invoke stoichiometry. For example, if $0.1\ \mathrm{mol}$ of the acid dissociated, it would certainly give $0.1\ \mathrm{mol}\ \ce{A-}$, but $0.1\ \mathrm{mol}$ would not be the amount of $[\ce{H3O+}]$ as it is part of another equilibrium. However, if we do make this approximation, ignoring the hydrolysis of $\ce{A-} $, we can justify it in some cases. Say if $K_\mathrm{a}$ is around $10^{-5}$ then the equilibrium constant for the hydrolysis would be $K_\mathrm{w}/K_\mathrm{b}$ which would be around $10^{-9}$. Thus it would not be a bad approximation to ignore the second equilibrium for finding the pH.