I'm going to assume for now that in the following figure three lines $BK$, $AL$ and $EE_1$ meet at one point $X$, and furthermore, $X$ lies on the circle $\Omega$. I'll prove it later.
Let $R_1$, $R_2$ and $R$ be the radii of $\Omega_1$, $\Omega_2$ and $\Omega$ respectively. I need them to introduce $k_1 = \frac{R}{R_1}$ and $k_2 = \frac{R}{R_2}$.
I'm going to prove now that $\frac{S_2}{S_1}$ depends only on $k_1$ and $k_2$ and nothing else.
$EE_1$ is radical axis of circles $\Omega_1$ and $\Omega_2$ so $pow(Y, \Omega_1) = pow(Y, \Omega_2) \Longrightarrow YK^2 = YL^2$.
Therefore $XY$ is median in $\triangle XKL$ and $EY$ is median in $\triangle EKL$.
So $S_{XKY} = S_{XLY}$ and $S_{EKY} = S_{ELY}$. Hence $S_{XKE} = S_{XLE}$. I'll name it $S$. So $S = S_{XKE} = S_{XLE}$.
Returning to what we want:
$$\frac{S_2}{S_1} = \frac{S_2}{S}\frac{S}{S_1} = \frac{BK}{KX}\cdot\frac{XL}{LA}.$$
Because of some basic similarities (or dilation, whichever you prefer)
$$1 + \frac{XL}{LA} = \frac{XA}{LA} = \frac{R}{R_1} = k_1,$$
so
$$\frac{XL}{LA} = k_1 - 1.$$
Similarly,
$$\frac{XK}{KB} = k_2 - 1.$$
Finally, $$\frac{S_2}{S_1} = \frac{BK}{KX}\cdot\frac{XL}{LA} = \frac{k_1 - 1}{k_2 - 1}.$$
The only thing left to do is to multiply 3 such fractions together, one for every pair of interior circles:
$$\frac{k_1 - 1}{k_2 - 1}\cdot\frac{k_2 - 1}{k_3 - 1}\cdot\frac{k_3 - 1}{k_1 - 1} = 1.$$
As promised, I'm going to show why $BK$, $AL$ and $EE_1$ meet on the big circle.
Consider the dilation with center A such that it transforms $\Omega_1 \rightarrow \Omega$.
Under that dilation point $L$ becomes some point $L'$ on the big circle. A dilation always transforms a line into a parallel line so tangent at $L'$ to the big circle is parallel to the line $KL$.
The very same thing happens to point $K$ under the dilation that transforms $\Omega_2 \rightarrow \Omega$. It becomes some point $K'$ on $\Omega$ and tangent line at $K'$ is also parallel to $KL$.
Fortunately, the number of points on $\Omega$ with that property is very limited, namely 1 (actually 2 but we are interested only in those to the other side of $KL$ than $A$ and $B$).
So considering that points inside each of the triples ($B$, $K$, $K'$) and ($A$, $L$, $L'$) are collinear, $BK$ and $AL$ meet on $\Omega$. Let's name that point $X$.
Great. One more line to go: $EE_1$.
So we want to prove that radical axis of $\Omega_1$ and $\Omega_2$ passes through $X$. In other words we want to prove that $pow(X, \Omega_1) = pow(X, \Omega_2)$ or $XK\cdot XB = XL\cdot XA$. So points $B$, $K$, $L$ and $A$ should be concyclic.
And they are. It's a simple matter of spotting equal angles so I'll just leave the following figure as illustration of where to look for them.
