Gauge condition in Maxwell's equation

Question

Maxwell's equations (differential forms formulation) read $$ dF = 0 \\ \partial^a F_{ab} = -j_b $$ where $j_b$ is the current-density 1-form. The first equation tells us there is some 1-form $A$ so that $F_{ab} = dA = \partial_a A_b - \partial_b A_a$. Then second equation then gives the following equation for $A$ itself: $$ \partial^a \partial_a A_b - \partial^a\partial_b A_a = \Box A_b - \partial_b \text{div}(A) = -j_b. \tag{1} $$ We note that $F = dA_f$ for any smooth function $f$, where $A_f := A + df$. We solve the equation $\Box f = -\text{div}A$ and see that (1) implies the following equation for $A_f$: $$ \Box (A_f)_b = -j_b \tag{2} $$ since $\Box (A_f)_b = \Box A_b + \partial_b \Box f$. My question is this. we then solve (2) instead of (1) since this is a nice wave equation. We then set $F = dA_f$. However, as (2) doesn't have any initial data prescribed, does this mean we can let $A_f$ be any solution to (2)? Secondly (and relatedly) do we need to choose a solution of (2) that has $\text{div}(A_f) = 0$ in order to get the right $F_{ab}$? Is such a condition a substitute for initial data for the wave equation?