Very nice problem. You add the other tangency point of each incircle and prove four lines concurrent instead (due to symmetry, this should be true). Of course, divide them into two pairs of three lines. The first pair is done by radical axis (need some angle chasing). The second pair uses a very clean menelaus. If anyone is interested, I’ll post the full solution.

Complex bash with $k_1$ as the unit circle. Let $k_1$ touch $BC$ at $P$, let $k_2$ touch $CD$ at $Q$, and let $k_1,k_2$ touch $AC$ at $T$. Then we have $$o_1 = 0$$$$|m|=|p|=|t|=1$$$$a=\frac{2mt}{m+t}$$$$b=\frac{2mp}{m+p}$$$$c=\frac{2pt}{p+t}$$Now $O_1,T,O_2$ are collinear. Thus let $\lambda$ be the radius of $k_2$, so that $\lambda < 1$ (since $AB < AD$) and $$o_2 = (1+\lambda)t$$If we let $z' = \frac{z-o_2}{\lambda}$, so that this transform sends $k_2$ to the unit circle, then we have $$t' = -t$$$$a' = -\frac{t\left[(\lambda-1)m + (\lambda+1)t\right]}{\lambda(m+t)}$$$$c' = -\frac{t\left[(\lambda-1)p + (\lambda+1)t\right]}{\lambda(p+t)}$$But we also have $$a' = \frac{2\ell't'}{\ell'+t'}$$$$c' = \frac{2q't'}{q'+t'}$$Matching the two equations for $a'$ gives $$-\frac{t\left[(\lambda-1)m + (\lambda+1)t\right]}{\lambda(m+t)} = -\frac{2\ell't}{\ell'-t}$$$$(\ell'-t)\left[(\lambda-1)m + (\lambda+1)t\right] = 2\lambda\ell'(m+t)$$$$\ell' = -\frac{t\left[(\lambda-1)m+(\lambda+1)t\right]}{(\lambda+1)m + (\lambda-1)t}$$and similarly, $$q' = -\frac{t\left[(\lambda-1)p+(\lambda+1)t\right]}{(\lambda+1)p + (\lambda-1)t}$$Shifting back gives $$\ell = \lambda \ell' + (1+\lambda)t = \frac{t\left[(3\lambda+1)m - (\lambda+1)t\right]}{(\lambda+1)m+(\lambda-1)t}$$$$q= \lambda q' + (1+\lambda)t = \frac{t\left[(3\lambda+1)p - (\lambda+1)t\right]}{(\lambda+1)p+(\lambda-1)t}$$We also have $$d' = \frac{2\ell'q'}{\ell'+q'} = -\frac{t\left[(\lambda-1)m+(\lambda+1)t\right]\left[(\lambda-1)p+(\lambda+1)t\right]}{(\lambda^2-1)mp + (\lambda^2+1)mt + (\lambda^2+1)pt + (\lambda^2-1)t^2}$$and so $$d = \lambda d' + (1+\lambda)t = \frac{t\left[(\lambda-1)(3\lambda+1)mp + (\lambda+1)^2mt + (\lambda+1)^2pt - (\lambda+1)^2t^2\right]}{(\lambda^2-1)mp + (\lambda^2+1)mt + (\lambda^2+1)pt + (\lambda^2-1)t^2}$$Now let $X$ be the intersection of lines $LM$ and $O_1O_2$. Since $X$ lies on $O_1O_2$, we have $$\overline{x} = \frac{x}{t^2}$$Meanwhile, we find the vector $$m-\ell = \frac{(\lambda+1)(m-t)^2}{(\lambda+1)m + (\lambda-1)t}$$Now $$\frac{m-x}{m-\ell} \in \mathbb{R}$$so $$\frac{(m-x)\left[(\lambda+1)m+(\lambda-1)t\right]}{(\lambda+1)(m-t)^2} = \frac{(t^2-mx)\left[(\lambda-1)m+(\lambda+1)t\right]}{t(\lambda+1)(m-t)^3}$$$$t(m-x)\left[(\lambda+1)m + (\lambda-1)t\right] = (t^2-mx)\left[(\lambda-1)m + (\lambda+1)t\right]$$$$x = \frac{t^2\left[(\lambda-1)m + (\lambda+1)t\right] - tm\left[(\lambda+1)m + (\lambda-1)t\right]}{m\left[(\lambda-1)m+(\lambda+1)t\right] - t\left[(\lambda+1)m + (\lambda-1)t\right]} = -\left(\frac{\lambda+1}{\lambda-1}\right)t$$(Since $\lambda < 1$, this denominator is nonzero.) It remains to show that $B, D, X$ are collinear. Now we have the vectors $$b-x = \frac{2(\lambda-1)mp + (\lambda+1)mt + (\lambda+1)pt}{(\lambda-1)(m+p)}$$$$d-x = \frac{2\lambda^2t\left[2(\lambda-1)mp + (\lambda+1)mt + (\lambda+1)pt\right]}{(\lambda-1)\left[(\lambda^2-1)mp + (\lambda^2+1)mt + (\lambda^2+1)pt + (\lambda^2-1)t^2\right]}$$Thus dividing gives $$\frac{b-x}{d-x} = \frac{(\lambda^2-1)mp + (\lambda^2+1)mt + (\lambda^2+1)pt + (\lambda^2-1)t^2}{2\lambda^2t(m+p)}$$which is real. $\blacksquare$

I claim that all three line intersect at the center of positive homothety that translates $k_1$ into $k_2$(call it $N$). Obviously, $O_1O_2$ passes through it. Now if $k_1$ and $k_2$ are tangent at $F$, then $AF=AM=AL$, so there exists a circle(call it $\omega_1$, that is tangent to $AB$ at $M$ and to $AD$ at $L$. Using three homotheties theorem on $k_1, k_2, \omega_1$, we obtain that $ML$ passes through $N$ Now $AM=AL=\frac{AB+AC-BC}{2}=\frac{AD+AC-CD}{2}$, from which we get that $AB+CD=BC+AD$. Then we can inscribe a circle into quadraliteral $ABCD$. Using three homotheties theorem on $k_1, k_2$ and the incircle, we obtain that $BD$ passes through $N$.

Let $k_1$ and $k_2$ be tangent at $T$ and let $k_1$ and $k_2$ touch $CB$ and $CD$ at $N$ and $K$, respectively. Notice that $ABCD$ must have an incircle. Claim: $LM$, $KN$, and $O_1O_2$ concur Notice that $A$ and $C$ are the circumcenters of $(LTM)$ and $(KTN)$ giving that these two circles are tangent to $O_1O_2$. Notice that $KLMN$ is cyclic as it has sides parallel to the in touch quadrilateral of $ABCD$. The result follows by R.A. on $(LTM)$, $(KTN)$, and $(KLMN)$. Claim: $LM$, $KN$, and $BD$ concur This follows by Menalus as $$\frac{AL}{LD}\cdot \frac{BM}{MA}=\frac{CK}{KD}\cdot\frac{BN}{NC}$$

First notice that $ABCD$ must have an incircle. Let the incircle intersect the sides $BA, AD, DC, CB$ at $E,F,G,H$ respectively. Let $T$ be the common tangency point of $k_1$ and $k_2$, and $K$ and $N$ be the other intouch points of $k_1$ and $k_2$. Let $X$ be the exsimilicenter of $k_1$ and $k_2$. It is clear that $T, O_1, O_2$ passes through $X$. Furthermore, $EFGH$ and $MLKN$ are homothetic so, $KLMN$ is cyclic. So, $LM$ and $KN$ must pass through $X$ as well. We now show that $X$ lies on $BD$. Indeed, it is known that $EF$, $GH$ and $BD$ concour (say, by Menelaus) and let this point be $Y$. Now as $\triangle LXK$ and $\triangle FYG$ are homothetic, we get that $Y,X,D$ collinear. So, $X$ lies on $BD$ as needed.