= Solution
Define the signed <dimensionless> displacement
$$
\lambda=\sqrt{\frac{m\omega}{2\hbar}}x_0.
$$
The <spatial translation operator> is $T(x_0)=e^{-ipx_0/\hbar}$, and the phases of the new <energy eigenstates> are fixed by $|n'\rangle=T(x_0)|n\rangle$. Their expansion coefficients are therefore
$$
\langle n'|0\rangle=\langle n|T(-x_0)|0\rangle.
$$
Since $p=i\sqrt{m\hbar\omega/2}(a^\dagger-a)$,
$$
T(-x_0)=e^{-\lambda(a^\dagger-a)}.
$$
Set $A=-\lambda a^\dagger$ and $B=\lambda a$. Their <commutator> is $[A,B]=\lambda^2I$, which commutes with both operators. Thus the stated special case of the <Baker--Campbell--Hausdorff formula> is applicable and gives
$$
e^{A+B}=e^{-\lambda^2/2}e^Ae^B.
$$
Because $a|0\rangle=0$,
$$
T(-x_0)|0\rangle=e^{-\lambda^2/2}e^{-\lambda a^\dagger}|0\rangle
=e^{-\lambda^2/2}\sum_{n=0}^\infty\frac{(-\lambda)^n}{\sqrt{n!}}|n\rangle.
$$
Taking the overlaps with $\langle n|$ therefore proves
$$
\boxed{|0\rangle=e^{-\lambda^2/2}\sum_{n=0}^\infty\frac{(-\lambda)^n}{\sqrt{n!}}|n'\rangle}.
$$
This is the <coherent state after shifting an oscillator equilibrium>. Its normalization follows from $e^{-\lambda^2}\sum_n\lambda^{2n}/n!=1$.
There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hbar))^{n/2}=|\lambda|^n$ agrees with $(-\lambda)^n$ when $x_0\leq0$, equivalently $\gamma\geq0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.
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