First, we try to determine the general form of a rotated ellipse in the first quadrant such that it remains tangent to the axes. We suppose it has the parametric equation
$$(x(t),y(t)) = (4 \cos u \cos t - 2 \sin u \sin t + h, 2 \cos u \sin t + 4 \sin u \cos t + k), \quad t \in [0,2\pi)$$ where $(h,k)$ is the center, and $u$ is the counterclockwise rotation angle of the ellipse relative to the coordinate axes. (Note I have modified the direction of rotation compared to the linked answer.)
Such an ellipse has horizontal tangent lines satisfying $$0 = \frac{dy}{dt} = 2 \cos u \cos t - 4 \sin u \sin t,$$ or $$t_{\text{crit}} = \arctan \frac{\cot u}{2}.$$ For these values of $t_{\text{crit}}$, we need to find $k$ such that $y(t_{\text{crit}}) = 0$, placing this ellipse so it is tangent to the $x$-axis; i.e, $$k = 2 \sqrt{\cos^2 u + 4 \sin^2 u}.$$ This gives, as a function of the angle of rotation $u$, the necessary vertical translation to make the ellipse tangent. A similar process using $dx/dt$ gives the necessary horizontal translation, which we show without proof: $$h = 2 \sqrt{4 \cos^2 u + \sin^2 u}.$$ Thus our ellipse is fully parametrized.
The locus of the center is simply $(h,k)$ as a function of $u$:
$$(h(u), k(u)) = \left(2 \sqrt{4 \cos^2 u + \sin^2 u}, 2 \sqrt{\cos^2 u + 4 \sin^2 u}\right).$$
A short computation of $h^2 + k^2$ shows that this locus is an arc of a circle, not the complete circle.
Where are the foci? We can first observe that they are located at some point along the line joining $(x(0), y(0))$ and $(x(\pi), y(\pi))$, so they have coordinates of the form $$(1-\lambda)(x(0), y(0)) + \lambda (x(\pi), y(\pi))$$ for $$\lambda = \frac{4 + 2 \sqrt{3}}{8} = \frac{2 + \sqrt{3}}{4}, \quad \text{and} \quad \lambda = \frac{2 - \sqrt{3}}{4}.$$ We again skip the calculation and show the result: $$(x_f(u), y_f(u)) = 2 \left(\sqrt{3} \cos u + \sqrt{4 \cos^2 u + \sin^2 u}, \sqrt{3} \sin u + \sqrt{\cos^2 u + 4 \sin^2 u} \right). \tag{1}$$ Note that this curve gives the locus of both foci, where $u$ and $u + \pi$ represent the location of each focus for a given rotation angle $u$.
All put together, we can visualize these loci in the following animation:
The conversion of the parametric formula to an implicit curve is tedious but not intractable; one would start with computing the square, then show that the square of the locus satisfies $$(x + y)(16 + xy) = 64 xy.$$
One approach to converting the locus is to note that we can write $$(x_f(u), y_f(u)) = \left(2 \sqrt{3} \cos u + \sqrt{(2 \sqrt{3} \cos u)^2 + 4}, 2 \sqrt{3} \sin u + \sqrt{(2 \sqrt{3} \sin u)^2 + 4} \right),$$ therefore $x_f(u)$ and $y_f(u)$ are roots of the quadratics $$x^2 - (4 \sqrt{3} \cos u) x - 4 = 0, \\ y^2 - (4 \sqrt{3} \sin u) y - 4 = 0,$$ or equivalently, $$48 \cos^2 u = \frac{(4-x^2)^2}{x^2}, \quad 48 \sin^2 u = \frac{(4-y^2)^2}{y^2}.$$ Thus $$48 = \frac{(4-x^2)^2}{x^2} + \frac{(4-y^2)^2}{y^2},$$ and the rest is an exercise in algebra.
