Yes, this works. Let me write $\mu$ and $\nu$ for the (signed) measures induced by $u$ and $v$, respectively. I then claim that $$ \mu(I\cap A)=\nu(I\cap A) \quad\quad\quad\quad (1) $$ for any open interval $I$. We can assume here that points of $A$ accumulate at both endpoints of $I$; otherwise, we simply make $I$ smaller. Since $\mu((a,x))=\nu((a,x))$ at all $x\in A$, (1) now follows from dominated convergence.

By regularity, (1) implies that $\mu|_A=\nu|_A$, and thus the total variations agree also.

Yes, this works. Let me write $\mu$ and $\nu$ for the (signed) measures induced by $u$ and $v$, respectively. I then claim that $\mu|_A=\nu|_A$, and this will then imply that the total variations agree also.

To prove my claim, it suffices to discuss $\mu(U), \nu(U)$ for open $U\supseteq B$, where $B$ is a Borel subset of $A$ (by regularity). Let's focus on a component $I=(c,d)$ of $U$. Here we can assume that $c,d$ are in $A$ or there are sequences $a_n\in (c,d)\cap A$ that converge to the endpoints; if that is not the case, then we can simply make $I$ smaller. 

Since $\mu((a,x))=\nu((a,x))$ for all $x\in A$ by the definition of $A$, we now obtain that also $\mu(I)=\nu(I)$, by approximation (use dominated convergence).