= Solution
The stated annihilation conditions can be solved algebraically, even though (b) shows they are not ground-state conditions for the full printed <Hamiltonian>. Invert the <canonical transformation>:
$$
A=a\cosh\theta+b^\dagger\sinh\theta,\qquad
B=b\cosh\theta+a^\dagger\sinh\theta.
$$
For $|\psi_\lambda\rangle=\mathcal N e^{\lambda a^\dagger b^\dagger}|0,0\rangle$, the <commutators> give
$$
a|\psi_\lambda\rangle=\lambda b^\dagger|\psi_\lambda\rangle,\qquad
b|\psi_\lambda\rangle=\lambda a^\dagger|\psi_\lambda\rangle.
$$
Thus both $A$ and $B$ annihilate it precisely when
$$
\boxed{\lambda=-\tanh\theta.}
$$
The normalized <two-mode squeezed vacuum> is
$$
|\psi_\lambda\rangle=\sqrt{1-|\lambda|^2}
\sum_{n=0}^{\infty}\lambda^n|n\rangle_x|n\rangle_y,
\qquad |\lambda|<1.
$$
The factorial from the exponential cancels those generated by the two <creation operators>. Normalization follows from a <geometric series>; the problem does not require its value, but it makes the subsequent <partial trace> explicit.
For comparison, the actual ground wave function of the printed stable <Hamiltonian> is
$$
\psi_{\rm true}(x,y)=
\left(\frac{m^2\omega_+\omega_-}{\pi^2\hbar^2}\right)^{1/4}
\exp\!\left[-\frac m{4\hbar}
\left((\omega_++\omega_-)(x^2+y^2)+2(\omega_+-\omega_-)xy\right)\right].
$$
It follows directly from the two normal-mode Gaussian vacua. The proposed $|\psi_\lambda\rangle$ is instead the exact <ground state> of the distinct pair-coupling <Hamiltonian> $\hbar\omega(a^\dagger a+b^\dagger b+1)+g(ab+a^\dagger b^\dagger)$ under its stability condition. Omitting the remaining terms of $kxy$ silently would change the problem.
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