= Solution
On a common domain where all products are defined, expansion of the <operator commutator> gives
$$
A[B,C]+[A,C]B=ABC-ACB+ACB-CAB=\boxed{[AB,C]}.
$$
Use the usual quantum position and momentum operators with $p_j=-i\hbar\partial_j$, the <canonical commutation relation> $[x_i,p_j]=i\hbar\delta_{ij}$, and <wavefunctions> with the decay/domain conditions needed for adjoints. The cross-coordinate operators commute, so
$$
L^\dagger=p_yx-p_xy=xp_y-yp_x=L.
$$
For an eigenstate $L\psi=\lambda\psi$, hermiticity gives $\lambda\|\psi\|^2=\overline\lambda\|\psi\|^2$, hence its <eigenvalue> is real.
The product identity gives $[L,x]=i\hbar y$, $[L,y]=-i\hbar x$, $[L,p_x]=i\hbar p_y$, $[L,p_y]=-i\hbar p_x$. Thus the commutators with $x^2,y^2$ cancel, as do those with $p_x^2,p_y^2$, proving \b[$[L,H]=0$]. Each oscillator energy <eigenspace> is finite-dimensional and preserved by $L$; the Hermitian restriction of $L$ can be diagonalized within it. Therefore an energy eigenbasis can be chosen to diagonalize <angular momentum> as well.
For $\phi_0=e^{-(x^2+y^2)/(2\hbar)}$, differentiation gives $\Delta\phi_0=[(x^2+y^2)/\hbar^2-2/\hbar]\phi_0$, so $H\phi_0=\hbar\phi_0$. Also $[H,x]=-i\hbar p_x$ and $p_x\phi_0=ix\phi_0$, giving $H(x\phi_0)=2\hbar x\phi_0$; the same calculation holds for $y\phi_0$. Directly $L=-i\hbar(x\partial_y-y\partial_x)$, hence
$$
L\phi_0=0,\qquad L\phi_x=i\hbar\phi_y,\qquad L\phi_y=-i\hbar\phi_x.
$$
The simultaneous normalized states are
$$
\boxed{\psi_1=\frac{(x+iy)\phi_0}{\sqrt\pi\,\hbar},\qquad
\psi_2=\frac{(x-iy)\phi_0}{\sqrt\pi\,\hbar},\qquad
\lambda_1=\hbar,\quad\lambda_2=-\hbar,\quad E_1=E_2=2\hbar.}
$$
Their squared unnormalized <norm> is $\int r^2e^{-r^2/\hbar}\,d^2x=\pi\hbar^2$. Their <inner product> vanishes because the integral of $(x-iy)^2e^{-r^2/\hbar}$ is zero: the $xy$ term is odd and the $x^2,y^2$ terms cancel by rotational symmetry.
Put $a=e\mathcal E$. The new Hamiltonian has
$$
\boxed{[L,H']=-a[L,x]=-i\hbar ay.}
$$
For a nonzero applied coupling $a$, a common normalizable eigenstate would be killed by the commutator, so $y\psi=0$. It would have to be supported on the line $y=0$, a set of zero planar measure, and would be the zero vector in $L^2$. Thus there are no nonzero common eigenstates in this case. Merely having a nonzero commutator would not prove that conclusion for arbitrary operators; the multiplication operator here has trivial kernel. If $a=0$, the original simultaneous eigenstates remain.
For the <shifted two-dimensional oscillator in a uniform electric field>, complete the square:
$$
H'=\frac12(p_{x'}^2+p_{y'}^2+x'^2+y'^2)-\frac{a^2}2,\qquad x'=x-a,\quad y'=y.
$$
Therefore $\psi_1(x',y')$ and $\psi_2(x',y')$ have the same shifted energy
$$
\boxed{E'_1=E'_2=2\hbar-\frac{(e\mathcal E)^2}2.}
$$
The modified <angular momentum> is
$$
\boxed{L'=(x-e\mathcal E)p_y-yp_x=L-e\mathcal E p_y.}
$$
It is <angular momentum> about the displaced centre, commutes with $H'$, and has <eigenvalues> $\pm\hbar$ on the two translated states.
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