If you follow your original idea, but convert to spherical coordinates, it is possible to isolate the radial coordinate $r$.
Step 1:
$$ z = \frac{-n_{x}x-n_{y}y}{n_{z}}$$
Step 2:
$$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{n_{x}^{2}x^{2}+n_{y}^{2}y^{2}+2n_{x}n_{y}xy}{c^{2}n_{z}^{2}} = 1$$
Step 3 (convert to spherical coordinates):
$$x = r\sin\theta\cos\phi $$
$$y = r\sin\theta\sin\phi $$
$$z = r\cos\theta$$
After substitution, one can solve for r fairly easily:
$$r = \frac{1}{\sqrt{\frac{\sin^{2}\theta\cos^{2}\phi}{a^{2}}+\frac{\sin^{2}\theta\sin^{2}\phi}{b^{2}}+\frac{n_{x}^{2}\sin^{2}\theta\cos^{2}\phi+n_{y}^{2}\sin^{2}\theta\sin^{2}\phi+2n_{x}n_{y}\sin^{2}\theta\cos\phi\sin\phi}{n_{z}^{2}c^{2}}}} $$
The variables $n_{x}$,$n_{y}$ and $n_{z}$ can be found by creating two vectors in the plane of interest, and crossing those two vectors to get the normal vector (normal to the surface). Here's a nice little reminder of how to do that. Finding the scalar equation of a plane.
This doesn't exactly answer your question, but I hope it helps nonetheless. It would be wise to double-check my work, as there are lots of sines and cosines floating around.
Side Note**
You can also add an arbitrary offset to your plane by extending your initial equation of a plane to this:
$$ n_{x}x+n_{y}y+n_{z}z+d=0 $$
where $ d = -x_{0}n_{x}-y_{0}n_{y}-z_{0}n_{z}$.
But since the term $d$ has no radial dependence it is either very difficult (or impossible) to come up with an analytic expression. I've also attempted to add an offset to the ellipsoid instead of the plane, but it gave me similar difficulties. That said, this particular case could absolutely be solved numerically.
The family of ellipses handled in the quoted passage was chosen specifically to have a simple equation in polar coordinates. Indeed,
from the ratio
$$
\frac{r}{d-r\cos\phi}=e
$$
we easily get the polar equation
$$
r=\frac{de}{1+e\cos\phi}\tag{1}
$$
familiar to some of us from a course in celestial mechanics ;-)
Anyway, here the parameters that the user is free to choose are $d$ and $e\in[0,1)$.
The other relevant coefficients: $a,b,h,c$ are then functions of $d$ and $e$,
and cannot be chosen independently. For example, the center of an ellipse
in this family is at the point $C=(h,0)=(-de^2/(1-e^2),0)$. As one of the focal points is fixed at the origin, $F_1=(0,0)$, the other focal point is then
at $F_2=(2h,0)$, i.e. on the negative $x$-axis (assuming $d>0$).
So this family of ellipses does not include any of those with the familiar equation
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\tag{2}
$$
because the foci of those ellipses are off the origin unless $a=b$. But if $a=b$, then we have a circle, i.e. an ellipse with eccentricity $e=0$. But if $e=0$, equation $(1)$ immediately implies $r=0$, i.e. a circle with radius zero. So the only ellipse of form $(2)$ that is also of form $(1)$ is the degenerate ellipse consisting of a single point.
In other words. Here an ellipse is formed by the loci of the points $P$ with the property that the ratio of their distances from a focal point and from the directrix is a constant. On a circle the distance to a focal point is also a constant, so the distance to the directrix must also be a constant. For this to happen either the circle collapses to a single point, or the directrix is at infinite distance. The latter case can be gotten here by a limiting process: letting $d\to\infty$ in such a way that $de$ remains a constant $R$. Then we must have $e\to0$, and $(1)$ becomes $r=R$ as expected.
Yet in other words. If one of the ellipses under discussion has its center at the origin, then
$$
h=-\frac{de^2}{1-e^2}=0.
$$
This implies that either $d=0$ or $e=0$. In either case equation $(1)$ gives $r=0$, i.e. a single point.
Best Answer
Let $AB$, $CD$ be the principal axes of the ellipse, intersecting at center $O$, and let $P$ be any point on the ellipse, $H$ and $K$ its projections on the axes (see figure below). We have: $$ {OH^2\over OB^2}+{OK^2\over OD^2}=1. $$ Let now $A'$, $B'$, ... be the perpendicular projections of points $A$, $B$, ... on a given plane. As perpendicular projections preserve the ratios of segments on a line, we have: $$ {O'H'^2\over O'B'^2}+{O'K'^2\over O'D'^2}=1. $$ But this equation means that point $P'$ belongs to the ellipse having $A'B'$ and $C'D'$ as conjugate diameters. You can find the principal axes of that ellipse, for instance, following the method explained here.