I've tried (only graphically - program Geogebra) - I would say, the ellipsis would have the same focus.
Truth has user TonyK - there is an infinite family of ellipses that circumscribe the rectangle.
Edit - followed by:
$\frac{x^2}{a^2}+\frac{y^2}{16}=1,\quad (x=3, and \,y=2) \quad \Rightarrow a^2=12$
$\Rightarrow \frac{x^2}{12}+\frac{y^2}{16}=1$
This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just want to see how the angles are equal.)
$DEGF$ is the quadrilateral and $A$ and $B$ are the foci.
Using sine law in $\Delta DBE$ you get $\dfrac{DE}{BE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}$ and in $\Delta GBE$ you get $\dfrac{GE}{BE}=\dfrac{\sin\angle GBE}{\sin\angle BGE}$. Dividing the two you'll obtain: $\dfrac{DE}{GE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BGE}{\sin\angle GBE}\tag{i}$
Using sine law in $\Delta BDF$ you get $\dfrac{FD}{FB}=\dfrac{\sin\angle DBF}{\sin\angle BDF}$ and in $\Delta BGF$ you get $\dfrac{FG}{FB}=\dfrac{\sin\angle GBF}{\sin\angle BGF}$. Dividing the two you'll obtain: $\dfrac{FD}{FG}=\dfrac{\sin\angle DBF}{\sin\angle BDF}\cdot\dfrac{\sin\angle BGF}{\sin\angle GBF}\tag{ii}$
Dividing (i) and (ii):$\dfrac{DE\cdot FG}{GE\cdot FD}=\dfrac{\sin\angle BGE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}\tag{iii}$
Using sine law in $\Delta DFI$ you get $\dfrac{DI}{FI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}$ and in $\Delta GFI$ you get $\dfrac{GI}{FI}=\dfrac{\sin\angle GFE}{\sin\angle FGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}\tag{iv}$
Using sine law in $\Delta DEI$ you get $\dfrac{DI}{EI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}$ and in $\Delta GEI$ you get $\dfrac{GI}{EI}=\dfrac{\sin\angle GEF}{\sin\angle EGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{v}$
Combining (iv) and (v): $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{vi}$
Doing the same thing with $DB$ and $GB$ you will get $$\dfrac{DB}{GB}=\dfrac{\sin\angle DFE}{\sin\angle GFE}\dfrac{\sin\angle BGF}{\sin\angle BDF}=\dfrac{\sin\angle DEF}{\sin\angle GEF}\dfrac{\sin\angle BGE}{\sin\angle BDE}\tag{vii}$$
Dividing (vi) and (vii) you get:$$\dfrac{\sin\angle FGD}{\sin\angle FDG}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}=\dfrac{\sin\angle DGE}{\sin\angle GDE}\cdot\dfrac{\sin\angle BDE}{\sin\angle BGE}\tag{viii}$$
Now due to the Property 1 on this link, $\angle FGD=\angle BGE, \angle GDE=\angle BDF,\angle DGE=\angle BGF$ and $\angle BDE=\angle FDG$. Making the necessary substitutions, relation (viii) becomes:$$\sin^2\angle BGE \cdot\sin^2\angle BDF=\sin^2\angle BDE \cdot\sin^2\angle BGF$$ $$\sin\angle BGE \cdot\sin\angle BDF=\sin\angle BDE \cdot\sin\angle BGF\tag{ix}$$
Combining (iii) and (ix): $DE\cdot FG=GE\cdot FD$
NOTE: Making use of the relations (vi),(vii) and (ix) it is easy to see that $\dfrac{DI}{GI}=\dfrac{DB}{GB}$. By the Angle Bisector Theorem, this means that $\angle DBF = \angle GBF$ and similarly $\angle FAG = \angle EAG$.
Best Answer
There is unique inscribed ellipse of a convex pentagon (dual case for $5$ points defining a conic). There are one and two degrees of freedom of drawing an inscribed ellipse in a (convex) quadrilateral and triangle respectively.
By means of skew transformation, we can transform an irregular quarilateral (convex but not parallelogram) into one with one pair of opposite sides are perpendicular.
$$(x',y')=(x+y\cos \omega,y\sin \omega)$$
Taking the vertices as $A(a,0)$, $B(b,0)$, $C(0,c)$ and $D(0,d)$ where $ab>0$, $cd>0$ and $(a-b)(d-c)>0$.
The two extreme cases are the ellipse degenerates into the diagonals.
Construct a family of conics touching with the axes with parameter $k$:
$$ \left[ k\left( \frac{x}{a}+\frac{y}{c} \right)+ (1-k)\left( \frac{x}{b}+\frac{y}{d} \right)-1 \right]^2=\lambda x y \tag{$\star$} $$
$$\lambda=4k(1-k) \left( \frac{1}{a}-\frac{1}{b} \right) \left( \frac{1}{d}-\frac{1}{c} \right)$$
$$4k(1-k) \left( \frac{1}{a}-\dfrac{1}{b} \right) \left( \frac{1}{d}-\dfrac{1}{c} \right) \left( \frac{k}{ac}+\frac{1-k}{bd} \right)>0 \implies k\in (0,1) $$
$$\text{centre}=\frac{ \left( \dfrac{k}{c}+\dfrac{1-k}{d},\dfrac{k}{a}+\dfrac{1-k}{b} \right)}{2 \left( \dfrac{k}{ac}+\dfrac{1-k}{bd} \right)}$$
See also another post of mine for the case of triangle here.
An illustration of a tangential quadrilateral. Note on the circular case at $k=0.6$:
Addendum
To generalize to any kind of convex quadrilateral, we may use the skew axes as the diagonals. Now taking the vertices as $A(a,0)$, $B(0,b)$, $C(c,0)$ and $D(0,d)$ where $ac<0$ and $bd<0$.
In tangential coordinates $(X,Y)$, tangent line $\frac{x}{a}+\frac{y}{b}=1$ can be written as $$Xx+Yy+1=0$$
Hence, the dual conic will pass through a "rectangle" with vectices $(-\frac{1}{a},-\frac{1}{b})$, $(-\frac{1}{c},-\frac{1}{b})$, $(-\frac{1}{c},-\frac{1}{d})$ and $(-\frac{1}{a},-\frac{1}{d})$, that is
$$\lambda (aX+1)(cX+1)+\mu (bY+1)(dY+1)=0$$
$$\det \begin{pmatrix} 0 & x & y & 1 \\ x & \lambda ac & 0 & \frac{\lambda (a+c)}{2} \\ y & 0 & \mu bd & \frac{\mu (b+d)}{2} \\ 1 &\frac{\lambda (a+c)}{2} & \frac{\mu (b+d)}{2} & \lambda+\mu \end{pmatrix}=0$$
The centre divides the Newton line, from $(0, \frac{b+d}{2})$ to $(\frac{a+c}{2},0)$ internally with ratio $\lambda:\mu$
Illustration of dual conics pair: