The relation $\sim$ is transitive. The argument is elementary, but rather tedious to spell out precisely, so I'll be light on the details.
Observe first of all that by decomposing further if necessary, we can get to a situation where all comparisons $\mu_j\leftrightarrow\nu_j$ are between two atoms $\mu_j=\nu_j=g\delta_n$ or between an atom and a measure supported by two points. This follows because I can decrease the number of points in the combined supports by decomposing further if I'm not in this situation yet. For example, if $$ a:=\max\textrm{supp}\,\mu_j>\max\textrm{supp}\,\nu_j=:b $$ and (let's say) $\nu_j(b)\le \mu_j(a)$, then I can represent $\nu_j(b)\delta_b$ as a convex combination of part $\mu_j$, and I have eliminated one point from the union of the supports. Similar arguments work in the other cases.
So if $\mu\sim\nu$, I can get from $\mu$ to $\nu$ by splitting and/or combining each point mass $\mu(n)\delta_n$, and for each such operation only two other points are involved. Also, since there are only finitely many other points within reach $A$, finitely many such operations suffice for a given $n$. In other words, I can partition each point mass $\mu(n)=w_1 + \ldots + w_{N_n}$ into finitely many parts, and each part $w_j$ gets a splitting or combining treatment.
Now if $\mu\sim\nu$, $\mu\sim\nu'$, then we can consider the common refinements of these partitions. We obtain corresponding finer decompositions of $\nu$, $\nu'$ (compared to the original ones), and these witness that $\nu\sim\nu'$.