I'm going to show that the product is
This is obvious if $D_1=D_2$, so we will assume WLOG $D_1<D_2$.
Let's start by making another drawing:
The 4 tangents intersect in points $B,C,B',C'$; there are two more intersection points which were discarded because they lie on the line through the circle centres. We have labelled the far of the two as $A$. Because the configuration is mirror symmetric the triangles $ABC$ and $AB'C'$ are congruent with sides $a=BC=B'C',b=CA=C'A,c=AB=AB'$ and the quadrilateral $BB'CC'$ is a trapezoid, in particular, it is cyclic; the circumcircle is indicated in orange. Let us call the sides of the trapezoid $x=CC',z=BB',y=BC'=CB'=c-b$.
Now it is time to call in the big boys
Ptolemy and Heron.
The desired product is by Ptolemy's theorem
(1) $xz = a^2-y^2 = (a+y)(a-y) = (a-b+c)(a+b-c)$.
The last product also occurs in Heron's formula for the area $S$ of $ABC$ (and $AB'C'$)
(2) $16S^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)$.
As the given circles happen to be the in- and one excircle of both $ABC$ and $AB'C'$ we can also express the area as
$4S=D_1(a+b+c)=D_2(-a+b+c)$ or multiplied together
(3) $16S^2 = (a+b+c)(-a+b+c)D_1D_2$.
Equating the r.h.s.s of (2) and (3), cancelling common factors and then comparing with (1) yields $xz = D_1D_2$ as contended.