Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(\pm c,0)$ for some $c\in[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=\sqrt{1-c^2}$. We must then prove that the equations
$$ \tag{1} x^2 + \left(\frac{y}{\sqrt{1-c^2}}\right)^2 = 1 $$
$$ \tag{2} \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$\tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $x\in(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
Best Answer
These are called $k$-ellipses. Yes, there is a literature. Here is one 2007 reference, which can lead you to others: "Semidefinite Representation of the $k$-Ellipse," arXiv:math/0702005v1:
![5-ellipse](https://i.stack.imgur.com/lhqta.jpg)
Caveat emptor: The Zariski closure of the 5-ellipse is an algebraic curve of degree 32(!). See their Fig.5: