Inversion with center $B$ and power $\overline{BC} \cdot \overline{BD}$ takes ${\omega_1}',$ ${\omega_2}'$ into themselves and carries $\omega_1,\omega_2$ into lines passing through the inverse $P$ of $A$ and perpendicular to ${\omega_1}',$ ${\omega_2}',$ due to conformity. In other words, ${PO_1}'$ and ${PO_2}'$ are perpendicular to $BO_1$ and $BO_2.$ Now, if $M,N$ are the intersections of $O_1O_2,$ ${O_1}'{O_2}'$ with $\ell$ and $L_{\infty}$ is the infinity point of $\perp \ell,$ then the pencils $B(O_1,O_2,M,L_{\infty})$ and $P({O_1}',{O_2}',L_{\infty},N)$ with corresponding perpendicular rays have equal cross ratios $\Longrightarrow$ $(O_1,O_2,M,L_{\infty})=({O_1}',{O_2}',L_{\infty},N)$ $\Longrightarrow$ ${MO_1:MO_2=N{O_2}':NO_1}'.$ Since $O_1O_2 \parallel {O_2}'{O_1}',$ then we deduce that $O_1{O_2}',$ $O_2{O_1}'$ and $MN \equiv \ell$ concur.

We don't need to use $L_\infty $. Here is a simplier way to end González's solution: Denote $\angle O_1'PN=\alpha, \angle O_2'PN=\beta$ and $X=O_1'P \cap O_2B, Y=O_2'P \cap O_1B$ then $\frac{O_1'N}{O_2'N}=\frac{\tan \beta}{\tan \alpha}=\frac{\tan XPB}{\tan YPB}=\frac{\cot XBP}{\cot YBP}=\frac{\tan O_1BM}{\tan O_2BM}=\frac{O_1M}{O_2N}$ Done