We start with a reformulation of a well-known statement on harmonic cyclic quadruples $(K_1, K_2, K_3, K_4)$, also provable by polar transformation (projective methods).
LEMMA. Being given four pairwise non-parallel lines $\ell_i$, $i = 1, 2, 3, 4$, tangent to a circle $\omega$ at points $K_i$, and such that lines $\ell_1, \ell_3$ and $K_2K_4$ are concurrent, then lines $\ell_2, \ell_4$ and $K_1K_3$ are also concurrent.

Proof. Let $O$ be the center of $\omega$, $X = \ell_1 \cap \ell_3 \cap K_2K_4$, $Y = \ell_2 \cap \ell_4$. We have $OX \perp K_1K_3$ and $OY \perp K_2K_4$. Let $Z = OX \cap K_1K_3$, $T = OY \cap K_2K_4$. Notice that triangles $\triangle OK_3X$ and $\triangle OZK_3$ are similar, and also similar are triangles $\triangle OK_4Y$ and $\triangle OTK_4$, hence $OY \cdot OT = OK_4^2 = OK_3^2 = OX \cdot OZ$.
This means that triangles $\triangle OXT$ and $\triangle OYZ$ are similar, hence $YZ \perp OX$, and so $Y \in K_1K_3$. $\square$

Suppose now lines $A_2A_3$, $A_4A_1$ and $T_1T_3$ are concurrent at a point $P$. Let $T'_4, T'_2$ be the tangency points of lines $A_4A_1$, respectively $A_2A_3$, to circle $\omega_1$, and let $T'_3$ be the second meeting point of line $T_1T_3$ and circle $\omega_1$. Let the tangent to $\omega_1$ at $T'_3$ meet the lines $A_4A_1$, $A_2A_3$ at points $A'_4$, respectively $A'_3$. The (direct) homothety of center $P$ that takes $\omega_1$ to $\omega_3$ maps $T'_3$ to $T_3$, hence $A_3A_4 \parallel A'_3A'_4$.
Let $Q = A_1A_2 \cap A_3A_4$, $Q' = A_1A_2 \cap A'_3A'_4$. Applying the LEMMA to circle $\omega_1$ and lines $A_2A_3$, $A'_3A'_4$, $A_4A_1$, $A_1A_2$, yields that points $Q'$, $T'_2$, $T'_4$ are collinear. The (inverse) homothety of center $A_1$ that takes $\triangle QA_1A_4$ to $\triangle Q'A_1A'_4$ maps $\omega_4$ to $\omega_1$, so maps $T_4$ to $T'_4$, hence $Q'T'_4 \parallel QT_4$. Similarly, the (inverse) homothety of center $A_2$ that takes $\triangle QA_2A_3$ to $\triangle Q'A_2A'_3$ maps $\omega_2$ to $\omega_1$, so maps $T_2$ to $T'_2$, hence also $Q'T'_2 \parallel QT_2$. Since points $Q'$, $T'_2$, $T'_4$ are collinear, it follows points $Q, T_2, T_4$ are also collinear.

The converse implication is done in a similar way, due to the cyclic nature of the notations used (just increase each index by 1).

Alternative Solution. (D. Şerbănescu) Suppose $Q, T_2, T_4$ are collinear. We will show $P, T_1, T_3$ are collinear. We will use the notations of the solution above, but also let $S'_1, S''_1$ be the tangency points of line $A_1A_2$ to circle $\omega_2$, respectively $\omega_4$, and let $S'_3, S''_3$ be the tangency points of line $A_3A_4$ to circle $\omega_2$, respectively $\omega_4$. Let $T''_4$ be the (other than $T_2$) meeting point of line $QT_2T_4$ and circle $\omega_2$, and let the tangent line to $\omega_2$ at $T''_4$ (parallel to $A_1A_4$) meet $A_2A_3$ at $P'$ (via the (direct) homothety of center $Q$ that takes $\omega_4$ to $\omega_2$).

Clearly $\triangle PT_2T_4 \sim \triangle P'T_2T_4''$ and $\triangle P'T_2T_4''$ is isosceles, so $PT_2 = PT_4$ (in other words, if $Q, T_2, T_4$ are collinear then $PT_2 = PT_4$; the other implication trivially also holds, but is irrelevant here).
From $PT_2 = PT_4$ and $PT'_2 = PT'_4$ follows $T'_2T_2 = T'_4T_4$. As external tangents, $T'_2T_2 = T_1S'_1$ and $T'_4T_4 = T_1S''_1$, hence $T_1$ is the midpoint of $S'_1S''_1$. Similarly, $T_3$ is the midpoint of $S'_3S''_3$. It follows that $P, T_1, T_3$ lie on the radical axis of the circles $\omega_2$ and $\omega_4$, hence are collinear.