Playing around with GeoGebra one finds that your approximation tends to overestimate the number of kissing ellipses, when $a/b$ grows. A possibly better formula can be found as follows.
Consider two consecutive tangent ellipses. Their major axes lie on two normal lines, meeting at some point $O$: the distance between $O$ and the tangency points is approximately equal to the radius of curvature $\rho$ of the ellipse at those points. The centers of those two ellipses are at an approximate distance of $2b$ between them and of $\rho+a$ from $O$. It follows that the distance between the tangency points is approximately $d=2b\rho/(a+\rho)$.
This distance of course varies along the central ellipse, but we can compute its average value $\overline{d}$:
$$
\overline{d}={1\over2\pi}\int_0^{2\pi}{2b\rho(t)\over a+\rho(t)}\,dt
$$
where $t$ is the usual parameter and
$$
\rho(t)={(a^2\sin^2t+b^2\cos^2t)^{3/2}\over ab}.
$$
By dividing the length of the ellipse by $\overline{d}$ we thus get an estimate for the number $N$ of kissing ellipses:
$$
N\approx{4 b E(1 - a^2/b^2)\over \overline{d}}.
$$
Unfortunately I didn't succeed in finding a simple expression for $\overline{d}$, which must then be evaluated numerically. In the graph below I plotted the value of $N$ given above as a function of $a/b$ (blue). For comparison I also plotted your result $N\approx 5a/b$ (red).
I think this is a better approximation than yours, but don't know how good it is.
EDIT.
I made some experimentation with GeoGebra: for $a/b=2.87$ my formula gives $N\approx13.3$, while I managed to put at most $12$ tangent ellipses tightly packed:
For $a/b=5.61$ my formula gives $N\approx21.8$, and $20$ tangent ellipses can be drawn (I'm showing two different arrangements):
It seems that the above formula for $N$, while still overestimating the actual number of ellipses, is quite good for large values of $a/b$.
EDIT 2.
Consider a point $C$ on the extension of the major axis of an ellipse, at a distance $R$ from its nearest vertex. A straightforward calculation shows that the angle $\varphi$ between the major axis and a tangent to the ellipse passing through $C$ is given by
$$
\varphi =\arctan{b\over\sqrt{R(R+2a)}}.
$$
Consider now two consecutive "kissing" ellipses (see diagram below), touching the inner ellipse at their vertices $P_1$ and $P_2$ and tangent between them at $T$. Let $C$ be the intersection of their (produced) major axes: line $CT$ is (to a very good approximation) a common tangent to both ellipses, forming angles $\varphi_1$ and $\varphi_2$ with their major axes, which can be computed with the above formula.
Distances $R_1$ and $R_2$ can be computed from the coordinates of $P_1$ and $P_2$, as well as the angle
$$
\angle P_1CP_2=\arccos{R_1^2+R_2^2-\overline{P_1P_2}^2\over2R_1R_2}.
$$
But this angle is the sum of $\varphi_1$ and $\varphi_2$, so we have the equality:
$$
\arccos{R_1^2+R_2^2-\overline{P_1P_2}^2\over2R_1R_2}=
\arctan{b\over\sqrt{R_1(R_1+2a)}}+\arctan{b\over\sqrt{R_2(R_2+2a)}}.
$$
From this equation one can, in principle, find $P_2$ once $P_1$ is given. This can be done numerically and can be used to build a routine which actually finds all kissing ellipses: starting with $P_1$ (which I took at one vertex of the inner ellipse) one can find $P_2$, then $P_3$ and so on, stopping with the last ellipse not intersecting the first one.
I wrote some Mathematica code which performs this task, thus computing the number of kissing ellipses. As an example, you can see below the case $a/c=20$, with $55$ tangent ellipses. The detail in the square is a magnification of the left vertex zone.
The plot below shows the number of kissing ellipses (black dots) as a function of $a/c$. The blue line is the approximate value found above.
For large values of $a/b$ I expect $N\approx 2a/b$, because a very thin ellipse should behave approximately like a $2a\times2b$ rectangle as $a/b\to\infty$, plus some tiny "border effect". This is confirmed by the numbers I got from Mathematica, which are plotted below:
I added for comparison the plot of my estimate (blue) and of $2a/b$ (red). As you can see, for very large values of $a/b$ the very simple rule $N\approx 2a/b$ is a better approximation. This trend is confirmed for even larger values:
$N(1000)=2064$, $N(2000)=4081$, $N(3000)=6094$, $N(4000)=8105$, $N(5000)=10128$.
As a last note, the equation for $\varphi$ above could be used to refine the rough estimate given at the beginning: a more precise approximation for the distance $d$ between two consecutive vertices is given by
$$
d=2\rho\arctan{b\over\sqrt{\rho(\rho+2a)}}.
$$
A quick test however shows that the estimate one gets from that is better than the older one for small values of $a/b$, but isn't a great improvement if $a/c$ is large.
