$a_n$-consistency implies weak consistency
This is only a partial answer to your full query, but I hope it gets you started by showing how you can connect the consistency property in question with the standard consistency property. Since $a_n |T_n - \theta| = O_p(1)$ there exists a function $c: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ such that:
$$\sup_{n \in \mathbb{N}} \mathbb{P}(a_n |T_n - \theta| \geqslant c(\varepsilon)) < \varepsilon
\quad \quad \quad \text{for all } \varepsilon > 0.$$
Consequently, for all $\varepsilon>0$ we have:
$$\sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big)
\leqslant \sup_{n \in \mathbb{N}} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) < \varepsilon.$$
Now, since $\{ a_n \}$ diverges to infinity, for any $\phi>0$ and $\epsilon>0$ you can choose $N$ sufficiently large to ensure that $a_n \geqslant c(\varepsilon)/\phi$ for all $n \geqslant N$. Consequently, for any $\phi>0$ and $\epsilon>0$ you have:
$$\begin{align}
\lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi)
&\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P}(|T_n - \theta| > \phi) \\[6pt]
&= \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| > \frac{\phi a_n}{a_n} \Big) \\[6pt]
&\leqslant \lim_{N \rightarrow \infty} \sup_{n \geqslant N} \mathbb{P} \Big( |T_n - \theta| \geqslant \frac{c(\varepsilon)}{a_n} \Big) \\[8pt]
&< \lim_{N \rightarrow \infty} \varepsilon \\[8pt]
&= \varepsilon. \\[6pt]
\end{align}$$
Since this inequality holds for all $\varepsilon>0$, we can say that for all $\phi>0$ we have:
$$\lim_{n \rightarrow \infty} \mathbb{P}(|T_n - \theta| > \phi) = 0,$$
which demonstrates weak consistency of $T_n$ in estimating $\theta$.
