We consider the configuration in which $Y$ lies between $A$ and $X$; the other configurations are treated analogously. Let $C$ be the intersection of $\ell_2$ and $\Gamma_1$. Then $C$ is the reflection of $A$ in $O_1O_2$. We get
$$\begin{align*} \angle O_1YX &= 180^\circ - \angle O_1YA && \text{(straight angle)} \\ &= 180^\circ - \angle YAO_1 && \text{($O_1YA$ is isosceles)} \\ &= 180^\circ - \angle XAO_1 && \text{($A$ is the reflection of $C$ in $O_1X$)} \\ &= 180^\circ - \angle XCO_1 && \text{($A$ is the reflection of $C$ in $O_1X$)} \end{align*}$$
which yields that $O_1CXY$ is cyclic.

Now let $S$ be the intersection of the line through $Y$ tangent to $\Gamma_1$, and the line $\ell_2$. Then both $SC$ and $SY$ are tangent to $\Gamma_1$, hence we have $\angle SCO_1 = 90^\circ = \angle SYO_1$, and $O_1CSY$ is cyclic.

We see that both $X$ and $S$ lie on the circle through $O_1$, $C$, and $Y$. Therefore, we have $\angle SXO_1 = \angle SYO_1 = 90^\circ$. We conclude that $SX$ is perpendicular to $O_1O_2$. Analogously, for the intersection $S'$ of the line through $Z$ tangent to $\Gamma_2$, and the line $\ell_2$, we can deduce that $S'X$ is perpendicular to $O_1O_2$. Because $S$ and $S'$ both lie on $\ell_2$, we have $S = S'$. Hence the two tangents intersect each other on $\ell_2$. $\square$