Solutions of the Laplace equation $\Delta u = 0$ include harmonic polynomials, which grow at most polynomially, each $|u(x)| \le C |x|^N$ for some $N>0$ depending on $u$. Define $D[v] = e^{-g(x)} \Delta (e^{g(x)} v)$ as an elliptic operator on $\mathbb{R}^n$. Its solutions include just $v(x) = e^{-g(x)} u(x)$ for any harmonic polynomial $u(x)$. Since $g(x)$ can grow arbitrarily fast as $|x|\to \infty$, it can be chosen so that infinitely many solutions of $D[v] = 0$ are bounded, or even decay faster than $e^{-g(x)+\varepsilon|x|}$ for any $\varepsilon > 0$. So $D[v]$ is not Fredholm on any weighted function space that doesn't exclude these solutions.

The point of this observation is that it is probably hopeless that expect some standard set of weights in Lebesgue or Sobolev spaces are enough to make an arbitrary elliptic operator Fredholm on a non-compact domain. The question is probably without making some assumption about the asymptotic behavior of the coefficients of $D[v]$ at its non-compact ends.

Solutions of the Laplace equation $\Delta u = 0$ include harmonic polynomials, which grow at most polynomially, each $|u(x)| \le C |x|^N$ for some $N>0$ depending on $u$. Define $D[v] = e^{-g(x)} \Delta (e^{g(x)} v)$ as an elliptic operator on $\mathbb{R}^n$. Its solutions include just $v(x) = e^{-g(x)} u(x)$ for any harmonic polynomial $u(x)$. Since $g(x)$ can grow arbitrarily fast as $|x|\to \infty$, it can be chosen so that infinitely many solutions of $D[v] = 0$ are bounded, or even decay faster than $e^{-g(x)+\varepsilon|x|}$ for any $\varepsilon > 0$. So $D[v]$ is not Fredholm on any weighted function space that doesn't exclude these solutions.

The point of this observation is that it is probably hopeless that expect some standard set of weights in Lebesgue or Sobolev spaces are enough to make an arbitrary elliptic operator Fredholm on a non-compact domain. Likely, one can't avoid making some assumption about the asymptotic behavior of the coefficients of $D[v]$ at its non-compact ends.