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) \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)} \tag{2} \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}