Let $E_1$ be an ellipse fixed in the plane. Let $E_2$ be a second, possibly different ellipse, which rolls around without slippage outside $E_1$, touching perimeter-to-perimeter. Let $c_2(t)$ be the center of $E_2$ as a function of time $t$, where $t$ measures the progress of the rolling.
Q. Is the locus $c_2(t)$ ever an exact circle when it is not the case that both $E_1$ and $E_2$ are circles?
Image: Erik Mahieu Mathematica Demo.
It seems likely the answer is No, but perhaps one can cleverly "cancel out" two eccentricities...
(I presume the usual parametric equation $(a\cos t\quad b\sin t)^\top$ for the ellipse.)
In the general case of noncongruent ellipses, the incommensurate arclengths will certainly preclude a circular locus.
When the fixed and rolling ellipses are congruent, the locus of the center is the pedal curve of the ellipse with respect to its center, which is a quartic curve with the Cartesian equation, $(x^2 + y^2)^2 = 4 a^2 x^2 + 4 b^2 y^2$, and thus not a circle:
The locus is called the "elliptic lemniscate", or the lemniscate of Booth. (I am of course assuming that $a>b$; certainly, the center of a circle rolling on another circle will produce a circle!)
If you want to have a circle as a locus, then the tracing point should be the focus of the ellipse instead:
In particular, the circle thus generated is centered at $\left(-\sqrt{a^2 - b^2},0\right)$ and has a radius of $2a$.
(Mathematica code for generating the figures is available upon request.)