This is not a complete answer, but I think it has a chance to lead to one.
I think the best possible ellipse is unique. Suppose that $E_1$ and $E_2$ are distinct best possible ellipses. By an appropriate rescaling of everything in the directions of principal axes of $E_1$, without loss of generality $E_1$ is a round disk, say $D$, of radius $1/\sqrt\pi$.
Then the width of $E_2$ in some direction is the same as that of the round disk $E_1$ (that is, $2/\sqrt\pi$), where the width of a set in a given direction is defined as the width of the narrowest infinite band in this direction containing the set. This follows because the product of the widths of ellipse $E_2$ in the directions of its principal axes is $(2/\sqrt\pi)^2$.
So, there exist the following: (i) a real $t$; (ii) a vector $b\in\mathbb{R}^2$; (iii) an orthonormal basis $(e_1,e_2)$ of $\mathbb{R}^2$; and (iv) a (shearing) affine operator $A$ on $\mathbb{R}^2$ such that $Ae_1=b+e_1$, $Ae_2=b+e_2+te_1$, and $AE_1=AD=E_2$.
Let then $E_0:=\dfrac{I+A}2\,D$, where $I$ is the identity operator. Then $E_0$ is an ellipse of area $1$.
For real $y$, let $[u,v]=[u(y),v(y)]=K(y):=\{x\in\mathbb R\colon xe_1+ye_2\in K\}$ be the $y$-"cross-section" of $K$. Similarly define the $y$-"cross-sections" $E_1(y)=[r_1,s_1]$ and $E_2(y)=[r_2,s_2]$ of $E_1$ and $E_2$. Then the $y$-"cross-section" $E_0(y)$ of $E_0$ is $[r_0,s_0]=[\frac{r_1+r_2}2,\frac{s_1+s_2}2]$. Let $\oplus$ denote the symmetric difference, and let $|\cdot|$ denote the Lebesgue measure on $\mathbb{R}$ or $\mathbb{R}^2$. Then $|E_j\oplus K|=\int_{\mathbb{R}}\delta_j(y)\,dy$ for $j=0,1,2$, where $\delta_j(y):=|E_j(y)\oplus K(y)|$. So, if we could show that \begin{equation} \delta_0(y)\ge\tfrac12\,\delta_1(y)+\tfrac12\,\delta_2(y) \tag{*} \end{equation} and this inequality is strict for some $y$ given that $E_1$ and $E_2$ are distinct, then we would obtain the desired contradiction: $|E_0\oplus K|>\tfrac12\,|E_1\oplus K|+\tfrac12\,|E_2\oplus K|=|E_1\oplus K|=|E_2\oplus K|$.
Unfortunately, inequality $(*)$ does not hold in general, without any assumptions on the convex sets $E_j$. However, it will hold if we establish the following condition: for any optimal ellipse $E$ and any real $y$ we have either $E(y)\cap K(y)\ne\emptyset$ or $E(y)=\emptyset$ or $K(y)=\emptyset$.
Does this condition have to hold? If so, how to prove it?