Let $TP_n$ and $TQ_n$ denote the pushforward measures of $P_n$ and $Q_n$, respectively, induced by $T$. Observe that \begin{align*} d_{TV}(P_n,Q_n) \geq d_{TV}(TP_n,TQ_n) &\geq d_{TV}(P,Q) - d_{TV}(TP_n,P) - d_{TV}(TQ_n,Q) \\ &= 1 - d_{TV}(TP_n,P) - d_{TV}(TQ_n,Q), \end{align*} where the first inequality follows from the definition of $d_{TV}$ (as a supremum) and the second inequality follows from the "reverse triangle inequality" for norms. This shows that (*) holds if $T$ converges in total variation distance (since the last two terms would then vanish as $n\to \infty$).

With the conditions you have posed, each of the two inequalities can be tight, and there is no guarantee that the final terms converge to zero. Thus, heuristically speaking, we would not expect your condition to be sufficient. A proof of this would, of course, require a counter example.

Rather than assuming convergence in total variation, it would also suffice with at stronger notion of separability between $P$ and $Q$. The condition $d_{TV}(P,Q)=1$ is, in fact, equivalent to the existence of a measurable set $A$ such that $P(A)=1$ and $Q(A)=0$ (i.e., the supremum in the total variation distance is attained when it is equal to one). Suppose a stronger condition holds: There is an open set $A$ such that $P(A)=1$ and $Q(\overline{A})=0$. From the Portmanteau theorem, it follows from weak convergence that $\liminf_{n\to\infty}TP_n(A)\geq 1$ and that $\limsup_{n\to\infty}TQ_n(A)\leq \limsup_{n\to\infty}TQ_n(\overline{A})\leq 0$. We may subsequently conclude that these inequalities are equalities since probabilities are in [0,1]. Combined, we conclude that $$ \liminf_{n\to \infty} d_{TV}(P_n,Q_n) \geq \liminf_{n\to \infty} d_{TV}(TP_n,TQ_n) \geq \liminf_{n\to \infty}TP_n(A) - \limsup_{n\to \infty}TQ_n(A)=1. $$

Let $TP_n$ and $TQ_n$ denote the pushforward measures of $P_n$ and $Q_n$, respectively, induced by $T$. Observe that \begin{align*} d_{TV}(P_n,Q_n) \geq d_{TV}(TP_n,TQ_n) &\geq d_{TV}(P,Q) - d_{TV}(TP_n,P) - d_{TV}(TQ_n,Q) \\ &= 1 - d_{TV}(TP_n,P) - d_{TV}(TQ_n,Q), \end{align*} where the first inequality follows from the definition of $d_{TV}$ (as a supremum) and the second inequality follows from the "reverse triangle inequality" for norms. This shows that $(*)$ holds if $T$ converges in total variation distance (since the last two terms would then vanish as $n\to \infty$).

With the conditions you have posed, each of the two inequalities can be tight, and there is no guarantee that the final terms converge to zero. Thus, heuristically speaking, we would not expect your condition to be sufficient. A proof of this would, of course, require a counter example.

Rather than assuming convergence in total variation, it would also suffice with at stronger notion of separability between $P$ and $Q$. The condition $d_{TV}(P,Q)=1$ is, in fact, equivalent to the existence of a measurable set $A$ such that $P(A)=1$ and $Q(A)=0$ (in other words, the supremum in the total variation distance is attained). Suppose a stronger condition holds: There is an open set $A$ such that $P(A)=1$ and $Q(\overline{A})=0$. From the Portmanteau theorem, it follows from weak convergence that $\liminf_{n\to\infty}TP_n(A)\geq 1$ and that $\limsup_{n\to\infty}TQ_n(A)\leq \limsup_{n\to\infty}TQ_n(\overline{A})\leq 0$. We may subsequently conclude that these inequalities are equalities since probabilities are in [0,1]. Combined, we conclude that \begin{align*} \liminf_{n\to \infty} d_{TV}(P_n,Q_n) &\geq \liminf_{n\to \infty} d_{TV}(TP_n,TQ_n)\\ &\geq \liminf_{n\to \infty}TP_n(A) - \limsup_{n\to \infty}TQ_n(A)=1, \end{align*} which implies $(*)$.