Even though an answer has been given, I wanted to share my ansatz using complex analysis, under the assumption $h$ being entire.
Defining $$f(t) \equiv e^{\frac{\left(b(t-\omega)+\omega\right)^2}{2}} h(t)$$ the objective becomes $$0= PV\int_{-\infty}^\infty \frac{h(t+\omega)}{it}\, {\rm d}t \stackrel{x=t+\omega}{=} PV\int_{-\infty}^\infty \frac{h(x)}{i(x-\omega)}\, {\rm d}x \quad , \quad \forall \omega \in \mathbb{R} \, .$$ Now lets consider the contour integral $$0=\oint_{-\infty}^\infty \frac{h(z)}{i(z-\omega)} \, {\rm d}z = PV\int_{-\infty}^\infty \frac{h(x)}{i(x-\omega)}\, {\rm d}x - \pi h(\omega) + \lim_{R\rightarrow\infty} \int_0^\pi \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t $$ for an entire function $h(z)$. The LHS follows by Cauchys theorem and the integral is closed along the semi-circle in the upper half-plane. By the assumption the PV-integral vanishes and it follows $$\pi h(\omega) = \lim_{R\rightarrow\infty} \int_0^\pi \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \quad , \quad \forall \omega \in {\mathbb{R}} \, .$$ Expanding $h$ as a series and using the geometric series on the RHS $$\pi \sum_{n=0}^\infty c_n \omega^n = \lim_{R\rightarrow\infty} \int_0^\pi {\rm d}t \sum_{m,n=0}^\infty c_n R^{n-m} \omega^m e^{it(n-m)} \\ =\lim_{R\rightarrow \infty} \left( \pi \sum_{\substack{m,n=0 \\ n-m=0}}^\infty c_n \omega^m + \sum_{\substack{m,n=0 \\ n-m\neq 0 \text{ even}}}^\infty 0 + 2i \sum_{\substack{m,n=0 \\ n-m \text{ odd}}}^\infty \frac{c_n R^{n-m}}{n-m} \, \omega^m \right) \\ \stackrel{n-m=2k+1}=\pi \sum_{n=0}^\infty c_n \omega^n + 2i \lim_{R\rightarrow \infty} \sum_{m=0}^\infty \omega^m \sum_{k=0}^\infty \frac{c_{2k+1+m}R^{2k+1}}{2k+1} \quad , \quad \forall \omega \in {\mathbb{R}} \, .$$ Hence, canceling and comparing coefficients, it follows $$\lim_{R\rightarrow \infty} \sum_{k=0}^\infty \frac{c_{2k+1+m}R^{2k+1}}{2k+1} = 0 \quad , \quad \forall m\in {\mathbb{Z}_{\geq 0}} \, .$$ Since the limit must exist for all $m\geq 0$, this is only possible if $$c_k =0 \quad , \quad \forall k \in \mathbb{N}$$
Note however, that there is no condition on $c_0$. Therefore $h(t)=\text{ const.}$, depending on neither $t$ nor $\omega$.