Concerning homogeneous polynomials: Let $P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}$ be such a polynomial, of degree $n$ such that $C:=P^{-1}(\{c\})$ is a simple closed curve for all large enough $c>0$.
If $n$ is odd, then every line through the origin will have at most one point of intersection with $C$. So, then $C$ cannot be a simple closed curve for any real $c$ -- because every line through any point interior to a simple closed curve must intersect the curve in at least two points.
It remains to consider the case when $n$ is even. Then $C$ is symmetric about the origin, and hence so is the interior of $C$. Then the centroid of the interior is the origin, and it does not depend on the level $c$.
Consider now the case of an elliptic polynomial \begin{equation*} P(z)=P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}+\sum_{j=0}^{n-1}b_j x^j y^{n-1-j}+K|z|^{n-2}. \end{equation*} of (necessarily even) degree $n=2m$, where $z:=(x,y)$ and $K=O(1)$ (as $|z|\to\infty$). We shall assume that $P$ is convex outside a large enough disk centered at $z=0$ (I don't know if this holds for all elliptic polynomials).
It is not hard to see that for a polynomial $p(x)=ax^n+bx^{n-1}+\cdots$ with $a>0$ and even $n$, for all large enough $c>0$ the equation $p(x)=c$ has exactly two roots $x_-<x_+$, and (as $c\to\infty$)
\begin{equation*}
x_\pm=\pm\Big(\frac ca\Big)^{1/n}-(1+o(1))\frac b{na}. \tag{1}
\end{equation*}
Consider now the two opposite infinitesimal sectors of the interior (say $I(c)$) of the simple closed curve $C=P^{-1}(\{c\})$ between the rays $t$ and $t+dt$ and between the rays $t+\pi$ and $t+\pi+dt$, where $t$ is the polar angle in the interval $[0,\pi)$. By (1), the centroid of the union of these two sectors of $I(c)$ is at distance
\begin{equation*}
d(t)\sim \frac23\,\Big(r_+(t)\frac{|r_+(t)|^2}{|r_+(t)|^2+|r_-(t)|^2}
+r_-(t)\frac{|r_-(t)|^2}{|r_+(t)|^2+|r_-(t)|^2}\Big) \tag{2}
\end{equation*}
from the origin, where
\begin{equation*}
r_\pm(t)=\pm\Big(\frac c{a(t)}\Big)^{1/n}-(1+o(1))\frac{b(t)}{na(t)}
\end{equation*}
are the signed radii of these sectors, with
\begin{equation*}
a(t):=\sum_{j=0}^n a_j \cos^jt\, \sin^{n-j}t,\quad
b(t):=\sum_{j=0}^{n-1} b_j \cos^jt\, \sin^{n-1-j}t.
\end{equation*}
Simplifying (2), we get
\begin{equation*}
d(t)\sim\frac{2b(t)}{na(t)}.
\end{equation*}
Averaging now over all the pairs of opposite infinitesimal sectors, we see that the centroid converges to
\begin{align}
&\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\frac12\,\Big(\frac c{a(t)}\Big)^{2/n}
\Big/\int_0^\pi dt\,\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \\
&=\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\Big(\frac1{a(t)}\Big)^{2/n}
\Big/\int_0^\pi dt\,\Big(\frac1{a(t)}\Big)^{2/n}.
\end{align}