I'm not sure what you are getting wrong... Your expression, indeed, should go to 1 as β goes to 0. There must be some error in your implementation.
The Wigner transform of the canonical ensemble density, as you point out, may go negative, but not after integrating out the ps as introduced in (28) of Wigner's breathtaking 1932 paper, since it is just a conventional expectation value of positive semidefinite quantities. So, any indication of negativity is a pathology, not something chalked up to Wigner function idiosyncracies. Your expression is a marginal, a projection on the x axis after the Gaussian in p is integrated out to a β-dependent constant.
As a check of the picture, I'll summarize here the case for the quadratic potential, $x^2/2$, for which you have a closed expression for the Wigner function, (which, being a Gaussian, is positive semi-definite, a fortiori).
Let me non-dimensionalize out the useless $m=1=\omega$, absorbed in the variables, but maintain $\hbar$ visible. The oscillator harmonic hamiltonian then is just $(\hat p ^2 + \hat x ^2)/2$, with both x and p possessing dimensions of $\sqrt \hbar$, symmetrically--the convention of choice for phase-space!
In any decent phase-space quantization text (including our concise treatise) the full Wigner transform of the phase-space evolution operator for the oscillator is worked out. But, behold!, since the Bloch equation is but the Moyal evolution equation for $\beta=-it$, so $t=i\beta$, you automatically have the relevant complete, un-normalized Wigner function, the celebrated Mehler kernel,
$$
e_\star ^{-\beta H/\hbar}= \frac{1}{\cosh \beta/2} ~e^{-2 H \tanh(\beta/2) /\hbar}= \frac{1}{\cosh \beta/2} ~e^{- \tanh(\beta/2) \frac{x^2+p^2}{\hbar} } .
$$
Integrating out the p - dependence nets you the answer, a positive semidefinite probability density
$$
P(x) \propto e^{-\frac{x^2}{\hbar} ~ \tanh(\beta/2) }=
e^{-\beta \frac{x^2}{2\hbar} }~ e^{\frac{x^2}{\hbar} ~ \frac{\beta^3}{24} +... } = e^{-\beta \frac{x^2}{2\hbar} }~ \left (1 + \frac{x^2}{\hbar} ~ \frac{\beta^3}{24} +... \right ) ,
$$
where $\tanh (\beta/2)=\beta/2 ( 1-\beta^2/12 + ...) $. This, now, is manifestly positive semidefinite, as a bona-fide probability density should be.
To compare with your expression, you may reinstate the $\hbar \sqrt{m\omega}$ absorbed into x. But...What about the leading term $V''$? Well, it is only a constant, absorbable into the normalization above, discarded and ignored via the proportionality sign, and equivalent to multiplication by a term $(1-\beta^2 c+...)$ so properly relegated to higher orders. (After the momentum itegration above, the cosh in the denominator mutated to a square root of the sinh of the double argument.... all irrelevant and omitted.)
This is only to reassure you of the systematics of the structure, and to help catch errors. It is the reference point of any such calculation. If the nondimensionalization confused you instead of simplifying things, you may repeat all steps keeping the silly constants.