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 that contains the set and goes in 
the direction perpendicular to the given one. 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)\le\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 for all real $y$ such that $E_j(y)\cap K(y)\ne\emptyset\ \forall j\in\{1,2\}$ or $K(y)=\emptyset$. 

Anyway, we need $(*)$ to hold on the average, given that both ellipses $E_1=D$ and $E_2$ are optimal approximations of $K$; this "average" inequality seems likely.