Chain of circles internally tangent to an ellipse.
In the theory of Lucas sequences if $\,v\,$ is constant, and $$ x_{n+1} = v\,x_n - x_{n-1} \tag{1}$$ for all $\,n,\,(P=v,\;Q=1)\,$ then $$ x_n^2 - x_{n+1}x_{n-1} = u \tag{2}$$ for a constant $\,u.\,$ This implies that $$ x_n^2 -v\,x_n x_{n+1} + x_{n+1}^2 = u \tag{3}$$ for all $\,n\,$ because $$ x_{n-1}\!+\!x_{n+1} \!=\! v\,x_n, \;\; x_{n-1}x_{n+1}\!=\!x_n^2\!-\!u. \tag{4}$$ In your case, the constants are $$ u=4e^2b^2,\quad v=4e^2-2. \tag{5}$$ Also, check that if $$ r_k:=(x_{k}-x_{k-1})/2, \tag{6} $$ then $$ (r_n+r_{n+6})/r_{n+3} = v^3-3v \tag{7} $$ for all $\,n.$ This implies that $$ r_{n+6}(r_n + r_{n+6}) = r_{n+3}(r_{n+3}+r_{n+6}). \tag{8}$$ for all $\,n.\,$ By the way, there is a similar result for $\,(x_{n+m}+x_{n-m})/x_n\,$ and $\,(r_{n+m}+r_{n-m})/r_n\,$ for any integer $\,m.$
Note that my answer is completely based on equation $(3)$ which was given in the question. I have not used any of the geometric content of the question.
Let $P := (a \cos\phi, b \sin\phi)$ on an origin-centered ellipse with semi-radii $a$ and $b$; define $c := \sqrt{a^2-b^2}$, so that the ellipse's eccentricity is $e := c/a$. The line through $P$, normal to the ellipse —that is, in direction $(b\cos\phi,a\sin\phi)$— meets the $x$-axis at $K:= (k,0)$, where $k:= c^2/a \cos\phi$. So, $K$ is the center of a circle internally tangent to the ellipse at $P$, and its radius, $r$, is given by $$r^2 = |PK|^2 = \frac{b^2(a^2-c^2\cos^2\phi)}{a^2} = \frac{b^2(c^2-k^2)}{c^2} \tag{1}$$ so that $$\frac{r^2}{b^2}+\frac{k^2}{c^2}=1 \tag{2}$$ This allow us to write, for some $\theta$, $$r = b\sin\theta \qquad k = c \cos\theta \tag{3}$$
Now, suppose $\bigcirc K_0$ and $\bigcirc K_1$ are circles internally tangent to the ellipse, with respective centers and radii given by $(3)$ for $\theta = \theta_0$ and $\theta=\theta_1$. If these circles are tangent to each other (with $K_1$ "on the right" of $K_0$), then $$\begin{align} k_0 + r_0 &= k_1 - r_1 \\[4pt] \to\quad -2 c \sin\frac{\theta_0 + \theta_1}{2} \sin\frac{\theta_0 - \theta_1}{2} &= -2 b \sin\frac{\theta_0 + \theta_1}{2} \cos\frac{\theta_0 - \theta_1}{2} \\[6pt] \to\quad \tan\frac{\theta_0 - \theta_1}{2} &= \frac{b}{c} \\[6pt] \to\quad \theta_1 &= \theta_0 - 2\arctan\frac{b}{c} \\[6pt] &= \theta_0 - 2\arccos e \tag{4} \end{align}$$
More generally, if circles $\bigcirc K_i$, defined by $\theta = \theta_i$ in $(3)$, form a tangent chain, then $$\theta_i = \theta_0 - 2 i \arccos e \tag{5}$$ where index $i$ is subject to certain viability conditions (eg, $\theta_i \geq 0$) that we'll assume hold. Thus, defining $\varepsilon := 2\arccos e$, we have $$\begin{align} \frac{r_{i+j} + r_{i-j}}{r_i} &= \frac{b\sin(\theta_0-(i+j)\psi)+b\sin(\theta_0-(i-j)\varepsilon)}{b \sin(\theta_0-i\varepsilon)} \\[6pt] &= 2\cos j\varepsilon = 2\cos( 2j \arccos e ) \\[4pt] &= 2\,T_{2j}(e) \tag{6} \end{align}$$ where $T_{2j}$ is the $2j$-th Chebyshev polynomial of the first kind. Notably, the value of $(6)$ is independent of $i$. In particular, if we take $j=3$ and both $i=4$ and $i=7$, we can write $$\frac{r_{4-3}+r_{4+3}}{r_4} = 2\;T_{2\cdot 3}(e) =\frac{r_{7-3}+r_{7+3}}{r_7} \tag{7}$$ which gives the result. $\square$
Addendum. In this follow-up question, @g.kov asks when an ellipse allows a "perfect packing" of $n$ tangent circles along its axis. It seems reasonable to append here a justification of the condition given there.
In a perfect packing, the first and last circles in a chain are tangent to the ellipse at the endpoints of the axis, so that their radii match the ellipse's radius of curvature (namely, $b^2/a$) at those points. Thus, we have $$r_0 = r_{n-1} = \frac{b^2}{a} \quad\to\quad \sin\theta_0 = \sin\theta_{n-1} = \frac{b}{a} \quad\to\quad \cos\theta_0 = \cos\theta_{n-1} = e \tag{8}$$ We can say that $\theta_0 = \pi - \arccos e$ and $\theta_{n-1} = \arccos e$. By $(5)$, this implies $$\arccos e = \theta_{n-1} = \theta_0 - 2(n-1)\arccos e = (\pi - \arccos e) - 2(n-1)\arccos e \tag{9}$$ so that $$\pi = 2n\arccos e \qquad\to\qquad \cos \frac{\pi}{2n} = e \tag{10}$$ This is equivalent to @g.kov's condition for a perfectly-packable ellipse. $\square$