The operator (on $L^2(\mathbb R)$) has purely absolutely continuous spectrum $\sigma_{ac}=(-\infty,\infty)$ of multiplicity $2$. I don't think that's very easy to see, and it definitely depends on the specifics of the situation; for example, if we make the potential more negative still, then the operator will be in the limit circle case and the spectrum becomes discrete.

I'll give a sketch, but this will not be very detailed. I suggest to analyze the ODE $-y''-x^2y = Ey$ asymptotically. This is a standard procedure, but it will require some calculations. Let $Y=(y,y')^t$, so $Y'=\bigl(\begin{smallmatrix} 0 & 1 \\ E+x^2 & 0\end{smallmatrix}\bigr)Y$, and write $Y= TZ$, where $$ T=\begin{pmatrix} 1 & 1 \\ i\omega & - i\omega \end{pmatrix} , \quad\quad \omega = \sqrt{E+x^2}, $$ is chosen such that it diagonalizes the coefficient matrix. We find that $$ Z' = \begin{pmatrix} i\omega + g & -g \\ -g & -i\omega +g \end{pmatrix} Z, \quad\quad g(x) = \frac{\omega'}{2\omega} = \frac{x}{2(E+x^2)} . $$ Notice that this was a partial success: the new system is close to diagonal, and the whole procedure worked because $x^2$ was slowly varying in the sense that the derivative is much smaller than the function.

Next, write $Z=(1+Q)U$, so $U$ solves $$ Q'U + (1+Q)U' = i\omega \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}(1+Q)U + g\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} (1+Q)U . $$ We now choose $Q$ as a solution of $$ Q' = i\omega (DQ-QD) - g \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} Q, \quad\quad D= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} . $$ This can be solved explicitly, and we then see that we can in fact choose a solution $Q(x) = O(\omega^{-1/2})= O(x^{-1/2})$. This simplifies matters considerably if we are willing to ignore $L^1$ terms: $U$ solves $$ U'= \left[ \begin{pmatrix} i\omega + g & 0 \\ 0 &-i\omega + g \end{pmatrix} + R \right] U, \quad\quad R\in L^1 . $$ Now Levinson's theorem on the asymptotic integration of such systems gives us basis solutions of the form $U=e_j \omega^{1/2} e^{\pm i\alpha(x)}(1+o(1))$. We go back to $y$ and obtain two solutions of the form $y\simeq \omega(x)^{1/2}e^{\pm i\alpha(x)}$. Since all solutions are asymptotically of the same size, my claims about the spectrum follow (by the subordinacy theory).

The operator (on $L^2(\mathbb R)$) has purely absolutely continuous spectrum $\sigma_{ac}=(-\infty,\infty)$ of multiplicity $2$. I don't think that's very easy to see, and it definitely depends on the specifics of the situation; for example, if we make the potential more negative still, then the operator will be in the limit circle case and the spectrum becomes discrete.

I'll give a sketch, but this will not be very detailed. I suggest to analyze the ODE $-y''-x^2y = Ey$ asymptotically. This is a standard procedure, but it will require some calculations. Let $Y=(y,y')^t$, so $Y'=\bigl(\begin{smallmatrix} 0 & 1 \\ E+x^2 & 0\end{smallmatrix}\bigr)Y$, and write $Y= TZ$, where $$ T=\begin{pmatrix} 1 & 1 \\ i\omega & - i\omega \end{pmatrix} , \quad\quad \omega = \sqrt{E+x^2}, $$ is chosen such that it diagonalizes the coefficient matrix. We find that $$ Z' = \begin{pmatrix} i\omega + g & -g \\ -g & -i\omega +g \end{pmatrix} Z, \quad\quad g(x) = \frac{\omega'}{2\omega} = \frac{x}{2(E+x^2)} . $$ Next, write $Z=(1+Q)U$, so $U$ solves $$ Q'U + (1+Q)U' = i\omega \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}(1+Q)U + g\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} (1+Q)U . $$ We now choose $Q$ as a solution of $$ Q' = i\omega (DQ-QD) - g \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} Q, \quad\quad D= \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} . $$ This can be solved explicitly, and we then see that we can in fact choose a solution $Q(x) = O(\omega^{-1/2})= O(x^{-1/2})$. This simplifies matters considerably if we are willing to ignore $L^1$ terms: $U$ solves $$ U'= \left[ \begin{pmatrix} i\omega + g & 0 \\ 0 &-i\omega + g \end{pmatrix} + R \right] U, \quad\quad R\in L^1 . $$ Now Levinson's theorem on the asymptotic integration of such systems gives us basis solutions of the form $U=e_j \omega^{1/2} e^{\pm i\alpha(x)}(1+o(1))$. We go back to $y$ and obtain two solutions of the form $y\simeq \omega(x)^{1/2}e^{\pm i\alpha(x)}$. Since all solutions are asymptotically of the same size, my claims about the spectrum follow (by the subordinacy theory).