Let $(E_n)_n$ be any finite collection of centred ellipses in $\mathbb{R}^2$. Suppose that $E_n$ are pairwise non-homothetic (i.e. there is no positive constant $c>0$ such that $E_n = c E_m$). Now I look at the Minkowski sum $$K=\sum_{k=1}^n c_k E_k$$ where $c_k\in\mathbb{R}-\{0\}$. This again will give us a convex body $K\subset\mathbb{R}^2$.
Is there a reference for the fact that $K$ is not an ellipse (or a disk) again? Or is there an easy argument to witness this fact?
By "finite" you surely mean "finite of cardinality at least two".
There is indeed an easy argument.
A convex body $K$ is an origin-centered ellipse if and only if $\sup\{ x\cdot v, x \in K\}= \sqrt{ Q(v)}$ for $Q$ a positive-definite quadratic form on $\mathbb R^2$. So it's equivalent to say that we do not have an identity of the form
$$\sqrt{Q(v)} = \sum_{k=1}^n c_k \sqrt{ Q_k(v)} \textrm{ for all } v\in \mathbb R^2$$
where $n>1$ and $Q_1,\dots, Q_n$ are positive-define quadratic forms on $\mathbb R^2$, no $Q_i$ is equal to a scalar multiple of any $Q_j$ for $i\neq j$, and all $c_k>0$.
If $Q$ is a scalar multiple of one of the $Q_i$, such an identity would give a nontrivial linear relation beteen the $\sqrt{Q_1},\dots, \sqrt{Q_n}$, and if not, such an identity would give a nontrivial linear relation between $\sqrt{Q_1},\dots,\sqrt{Q_n},\sqrt{Q}$, which after increasing $n$ by $1$ and setting $Q_{n+1}=Q$ also gives a nontrivial linear relation between $\sqrt{Q_1},\dots, \sqrt{Q_n}$ where no $Q_i$ is equal to a scalar multiple of any $Q_j$ for $i\neq j$.
Such linear relations cannot exist.
An analytic proof is that the functions $\sqrt{Q_1},\dots, \sqrt{Q_n}$ each extend to multibranched holomorphic functions on $\mathbb C^2$ minus two lines where they are singular. When two quadratic forms are not scalar multiples of each other, their lines of singularity are distinct, so if we could write one of these square roots of forms as a linear combination of the others, we would write a function singular at a line as a linear combination of functions holomorphic on the line, which is impossible.
Algebraically, one can check that by induction in $m$ every function in the field $\mathbb R(x,y, \sqrt{ Q_1(x,y)},\dots, \sqrt{Q_m(x,y)})$ whose square is in $\mathbb R(x,y)$ is in fact an element of $\mathbb R(x,y)$ times a product of some of the $\sqrt{ Q_1(x,y)},\dots, \sqrt{Q_m(x,y)}$, and thus that $\sqrt{ Q_{m+1}(x,y)}$ is not in this field since $Q_{m+1}(x,y)$ can not be written as a perfect square times a product of some of $Q_1(x,y),\dots,Q_m(x,y)$, so in particular $\sqrt{Q_{m+1}(x,y)}$ is not a linear combination of $\sqrt{ Q_{m+1}(x,y)}, \sqrt{ Q_1(x,y)},\dots, \sqrt{Q_m(x,y)}$ and thus there can be no linear relations among $\sqrt{Q_1},\dots, \sqrt{Q_n}$.
