By Monge on $\omega_1$, $\omega_2$, and the circle passing through $A$ tangent to the line through $A$ parallel to $BC$ with radius $0$, we need to show $\frac{BQ}{QC}=\frac{BX}{CY}$, which follows from spiral similarity as $\triangle QBX\sim\triangle QCY$.

Let $b=-1$, $c=1$, $x=-1+s$, and $y=1+t$. We have $\frac{a+1}s,\frac{a-1}t\in\mathbb R$, so $\frac{\overline sa+\overline s-s}s=\overline a=\frac{\overline ta-\overline t+t}t$. Solving, $a=\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}$. We have $q=\frac{cx-by}{c+x-b-y}=\frac{s+t}{s-t}$. Also, letting $q_1=o_1+1$ and $q_2=o_2-1$, we have $(q_1-s)(\overline{q_1}-\overline s)=q_1\overline{q_1}$; substituting $\overline{q_1}=-q_1$ and solving gives $q_1=\frac{s\overline s}{\overline s-s}$. The exsimilicenter, which is the intersection of $O_1O_2$ and $BC$, is given by $z=\frac{o_1\overline{o_2}-\overline{o_1}o_2}{o_1-o_2-\overline{o_1}+\overline{o_2}}=\frac{(q_1-1)(-q_2+1)-(-q_1-1)(q_2+1)}{(q_1-1)-(q_2+1)-(-q_1-1)+(-q_2+1)}=\frac{2(q_1+q_2)}{2(q_1-q_2)}=\frac{s\overline s(\overline t-t)+t\overline t(\overline s-s)}{s\overline s(\overline t-t)-t\overline t(\overline s-s)}$. We are done because this equals the intersection of $AQ$ and $BC$, $\frac{a\overline q-\overline aq}{a-q-\overline a+\overline q}=\frac{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}\cdot\frac{\overline s+\overline t}{\overline s-\overline t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}\cdot\frac{s+t}{s-t}}{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}-\frac{s+t}{s-t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}+\frac{\overline s+\overline t}{\overline s-\overline t}}=\frac{(s-t)(2st-s\overline t-\overline st)(\overline s+\overline t)-(\overline s-\overline t)(-2\overline{st}+s\overline t+\overline st)(s+t)}{(s-t)(\overline s-\overline t)((2st-2s\overline t-2\overline st+2\overline{st}))+(\overline st-s\overline t)((s-t)(\overline s+\overline t)(\overline s-\overline t)(s+t))}=\frac{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)+t\overline t(\overline s-s))}{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)-t\overline t(\overline s-s))}=z$

Lemma: Let $A$, $B$, $C$, and $D$ be points on a circle, and let $\overline{AB}$ and $\overline{CD}$ intersect at $X$. Then, $\tfrac{CX}{XD}=\tfrac{CA}{AD} \cdot \tfrac{CB}{BD}$. Proof: We have \[\frac{CX}{XD}=\frac{[ABC]}{[ABD]}=\frac{\frac{1}{2} CA \cdot CB \cdot \sin \angle ACB}{\frac{1}{2} DA \cdot DB \cdot \sin \angle ADB}=\frac{CA}{AD} \cdot \frac{CB}{BD},\]as desired. $\square$ Let $\overline{AQ}$ and $\overline{BC}$ intersect at $Z$. Notice that $QBX \sim QCY$, so $\tfrac{QB}{QC}=\tfrac{BX}{CY}$. By the lemma, we have \[\frac{ZB}{ZC}=\frac{BA}{AC} \cdot \frac{BQ}{QC}=\frac{AB}{AC} \cdot \frac{BX}{CY}=\frac{AB}{AC} \cdot \frac{2BO_1 \cdot \cos \angle O_1BA}{2CO_2 \cdot \cos \angle O_2CA}=\frac{AB}{AC} \cdot \frac{BO_1}{CO_2} \cdot \frac{\sin \angle ABC}{\sin \angle ACB}=\frac{BO_1}{CO_2}.\]Thus, we have $ZBO_1 \sim ZCO_2$, so $Z$, $O_1$, and $O_2$ are collinear, as desired. $\square$ Remark: This problem has 3 main components: the circles and intersections, the definition of $Q$, and the definition of $Z$. These can be dealt with mostly independently of each other: using the lemma, length relations involving $Z$ decompose to those involving $Q$, which decompose by spiral similarity to those involving $X$ and $Y$, which are easily calculated.

Here's a solution using degree 1 moving points / spiral similarity. It is well-known that spiral similarities are linear maps. In fact, the following lemma is true. Lemma. (well-known) Let $l_1,l_2$ denote two lines intersecting in point $Z$. Let $O$ be a fixed point. If $X_t$ is a linearly moving point on $l_1$ and $Y_t=l_2\cap (AOX_t)$, then $\frac{X_0X_t}{Y_0Y_t}$ is constant. Proof. $O$ is the center of the spiral similarity sending $X_t$ to $Y_t$. Hence, $\triangle QX_0X_t\sim QY_0Y_t$, which means $\frac{X_0X_t}{Y_0Y_t}=\frac{QX_0}{QY_0}$, which is indeed a constant. $\blacksquare$ Fix point $Q$ on $(ABC)$. Choose a point $X$ line $AB$ and define $Y=(AQX)\cap AC$. Then, $X\rightarrow Y$ is a linear map (spiral similarity centered at $Q$). From our lemma, we may write $\frac{BX}{CY}=\frac{QB}{QC}$. Since $\omega_1,\omega_2$ are tangent to $BC$, we have $\measuredangle BO_1X=2\measuredangle CBX$ and $\measuredangle CO_2Y=2\measuredangle BCY$. As a result, we have \[\frac{O_1B}{O_2C}=\frac{BX}{CY}\cdot\frac{\sin(\measuredangle CBA)}{\sin(\measuredangle ABC)}=\frac{QB}{QC}\cdot\frac{AC}{AB}\]Let $T=AQ\cap BC$. For $T,O_1,O_2$ to be collinear, it suffices to show that \[\frac{TB}{TC}=\frac{O_1B}{O_2C}=\frac{QB}{QC}\cdot\frac{AC}{AB},\]which is a well-known identity (sketch: use sine rule in $\triangle TAB,\triangle TAC$). $\blacksquare$

Oh bruh why am I so washed at geometry again my solution is wayyy overcomplicated oops. So we start by defining $T=\overline{AQ}\cap\overline{BC}$. Note $|\operatorname{Pow}_\omega(T)|=TA\cdot TQ=TB\cdot TC=k^2$, so we can consider an inversion $I$ centred at $T$ with radius $\sqrt{k}$. Let $Y’=I(X)$ and $X’=I(Y)$ under this inversion. Note that $T-X-Y’$ and $T-X’-Y$. However I would like to show that $Y’\in\omega_2$ and $X’\in\omega_1$ for reasons to be disclosed later on. This is not hard; notice that $\omega_1\iff\omega_2$ under $I$ which is clear since they are both tangent to $\overline{BC}$ and $B\iff C$. Now I can disclose why I wanted $X’$ and $Y’$ to satisfy these properties; it’s quite clear now that under $I$ we must have $\omega_1\iff\omega_2$, so $\overline{O_1O_2}$ also passes through $T$. Edit: I just realised I proved wayyyy too much again (I proved that five lines concurred at $T$ instead of your normal three). To prove your normal three just take the inversion $I$ centred at $T$ with power $\operatorname{Pow}_\omega(T)$ and then note that $\omega_1\iff\omega_2$, done.