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 $|z|\to\infty$).
For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$. Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_-<x_+$, and \begin{equation*} x_\pm=\pm\Big(\frac c{d_n}\Big)^{1/n}-(1+o(1))\frac{d_{n-1}}{nd_n} \tag{1} \end{equation*} uniformly over all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ (as $c\to\infty$). This uniformity can be obtained by refining this reasoning.
The main idea for the elliptic polynomial case is to consider 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 (signed) 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)} \tag{2a} \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*} the asymptotic relations in (2) and (2a) are uniform in $t\in[0,\pi)$. Formula (2) follows because (i) the centroid of an infinitesimal sector of radius $r>0$ between the rays $t$ and $t+dt$ is at distance $\frac23\,r$ from the origin, (ii) the area of this sector is $\frac12\,r^2\,dt$, and (iii) the centroid of the union of the two sectors is the weighted average of the centroids of the two sectors, with weights adding to $1$ and proportional to the areas of the sectors, and thus proportional to the squared radii of the sectors.
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}. \tag{3}
\end{align*}
I have checked this result numerically for $P(x,y)=x^4 + y^4 + 3 (x - y)^4 + y^3 + x y^2 + 10 x^2$, getting the centroid $\approx(-0.182846, -0.245149)$ for $c=10^4$ and $\approx(-0.189242,-0.25)$ for the limit (as $c\to\infty$) given by (3). From the above reasoning, one can see that the distance of the centroid from its limit is $O(1/c^{1/n})$; so, the agreement in this numerical example should be considered good, better than expected.