Here's a brute force approach, with a possibly-unnecessary detour into complex arithmetic ...
Part 1: The Points
Let the ellipse have major radius $p$, aligned with the $x$ axis, and minor radius $q$, aligned with $y$-axis. Let its center lie at the origin. For any $\theta$, we can find a variable circle tangent to the major and minor auxiliary circles at respective points $P := (p\cos\theta, p\sin\theta)$ and $Q := (-q\cos\theta, -q\sin\theta)$; this circles center is $(P+Q)/2$, and its radius is $(p+q)/2$, so its equation works out to be
$$( 2 x - (p-q)\cos\theta)^2 + (2 y - (p-q)\sin\theta)^2 = ( a + b )^2 \tag{1}$$
We can parameterize a point on the ellipse as $E(\phi) := (a\cos\phi, b\sin\phi)$. Substituting into $(1)$ gives an implicit relation in $\phi$ for the points of intersection of the variable circle and the ellipse. With a little trigonometric massaging, this relation reduces to
$$(p-q)\; \sin\frac{\theta+\phi}{2}\;\left( p \cos\phi \sin\frac{\phi-\theta}{2} + q \sin\phi \cos\frac{\phi-\theta}{2} \right)=0 \tag{2}$$
where we'll reasonably ignore the first factor (so that we're restricting our attention to non-circular ellipses). Observe that
$$\sin\frac{\theta+\phi}{2}= 0 \tag{2a}$$
implies that $\theta+\phi$ is a multiple of $2\pi$; effectively $\phi = -\theta$. This happens to be the point that's left out of the concurrent-normals configuration. The points in the configuration correspond to three solutions of
$$p\cos\phi \sin\frac{\phi - \theta}{2}+ q\sin\phi\cos\frac{\phi-\theta}{2} = 0\tag{$\star$}$$
As-is, this equation may not look much like a cubic in $\phi$, but we can remedy that: dividing-through by the cosines gives
$$0 = p \tan\frac{\phi-\theta}{2} + q \tan 2 \frac{\phi}{2} = p \frac{\tan(\phi/2)-\tan(\theta/2)}{1+\tan(\phi/2)\tan(\theta/2)} + q \frac{2\tan(\phi/2)}{1-\tan^2(\phi/2)}$$
Defining $x_\star := \tan(x/2)$ to reduce clutter, the above becomes our obvious cubic:
$$p \phi_\star^3 - ( p + 2 q )\theta_\star \phi_\star^2 - (p + 2 q) \phi_\star + p \theta_\star = 0 \tag{$\star^\prime$}$$
Part 2: The Lines
Given a point $E(\phi) := ( p\cos\phi, q \sin\phi)$ on the ellipse, Calculus tells us that the line tangent to the ellipse at that point has direction vector $(-p\sin\phi, q\cos\phi)$, so that the normal line at that point has direction vector $n(\phi) := (q\cos\phi, p\sin\phi)$.
We can therefore parameterize the normals at points $A := E(\alpha)$, $B := E(\beta)$, $C := E(\gamma)$, by
$$A + a\,n(\alpha) \qquad,\qquad B + b\,n(\beta) \qquad,\qquad C + c\;n(\gamma)$$
Setting pairs of parameterizations equal, we can eliminate $a$, $b$, $c$ to reveal the condition under which the normals have a common point. Again, a little trigonometric massage reduces the condition to a tidy form:
$$\sin(\alpha+\beta) + \sin(\beta+\gamma) + \sin(\gamma+\alpha) = 0 \tag{$\star\star$}$$
A less-tidy form, but one that matches $(\star^\prime)$, comes from expanding $(\star\star)$ in terms of $\alpha_\star := \tan(\alpha/2)$, etc.
$$\alpha_\star + \beta_\star + \gamma_\star = \alpha_\star\beta_\star\gamma_\star\left( \alpha_\star \beta_\star + \beta_\star \gamma_\star + \gamma_\star \alpha_\star \right) \tag{$\star\star^\prime$}$$
Part 3. The Connections
To see that $(\star^\prime)$ and $(\star\star^\prime)$ actually complete the proof, recall Vieta's Formulas for the coefficients of a polynomial in terms of its roots. In particular, if $\alpha_\star$, $\beta_\star$, $\gamma_\star$ are the roots of $(\star^\prime)$, then
$$\begin{align}
\alpha_\star \beta_\star \gamma_\star &= -\frac{p}{p}\theta_\star = -\theta_\star \tag{3a} \\[4pt]
\alpha_\star \beta_\star + \beta_\star \gamma_\star + \gamma_\star \alpha_\star &= -\frac{p+2q}{p} \tag{3b} \\[4pt]
\alpha_\star + \beta_\star + \gamma_\star &= \phantom{-}\frac{p+2q}{p}\;\theta_\star\tag{3c}
\end{align}$$
Based on the right-hand sides, we have $(3c) = (3a)(3b)$; the corresponding equality for the left-hand sides is exactly $(\star\star^\prime)$. Thus, points that lie on the variable circle are points that determine concurrent normals. $\square$
