$\omega$-dependent solutions are not interesting, since after the substitution $t=x-\omega$ $$0=PV\int_{-\infty}^\infty e^{-\frac{\left(b(x-\omega)+\omega\right)^2}{2}} \frac{f(x)}{i(x-\omega)} \, {\rm d}x \, , \tag{1}$$ you could just set $$f(x) \equiv e^{\frac{\left(b(x-\omega)+\omega\right)^2}{2}} i(x-\omega) h(x)$$ and you would have $$0=PV\int_{-\infty}^\infty h(x) \, {\rm d}x$$ which certainly has many solutions for $h$. Therefore we can assume $f$ to not depend on $\omega$ and define $$f(x) \equiv e^{\frac{b^2x^2}{2}} h(x)$$ after which (1) simplifies to $$0=PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)} \, {\rm d}x$$ where the constant, $\omega$-dependent, factor was removed.
Now we assume $h$ to be entire and use contour integration to get $$0=\oint_{-\infty}^\infty e^{\omega z b(b-1)} \frac{h(z)}{i(z-\omega)} \, {\rm d}z \\ = PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)}\, {\rm d}x - \pi h(\omega) \, e^{\omega^2 b(b-1)} + \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \, .$$ The first line follows by Cauchys theorem and the integral is closed along the semi-circle in the upper half-plane. By the assumption of the vanishing PV-integral it follows $$\pi h(\omega) \, e^{\omega^2 b(b-1)} = \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \quad , \quad \forall \omega \in {\mathbb{R}} \, .$$
Now we expand both sides as a power-series in $\omega$. For the LHS we obtain $$\pi h(\omega) e^{\omega^2 b(b-1)} = \pi \sum_{n,m=0}^\infty c_n \omega^n \frac{\left(\omega^2b(b-1)\right)^m}{m!} = \pi \sum_{k=0}^\infty \omega^k \sum_{\substack{n,m=0 \\ n+2m=k}}^\infty \frac{c_n \left(b(b-1)\right)^m}{m!} \, .$$ Similarly the RHS becomes $$\lim_{R\rightarrow \infty} \sum_{n,m,k=0}^\infty c_n R^{n-k+m} \omega^{k+m} \frac{\left(b(b-1)\right)^m}{m!} \int_0^\pi e^{it(n-k+m)} \, {\rm d}t \\ =\lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k=0 \\ k+m=l}}^\infty c_n R^{n-k+m} \frac{\left(b(b-1)\right)^m}{m!} \\ \times \left( \pi \left[ n-k+m=0 \right] + 0 \left[n-k+m | \text{even} \right] + \frac{2i}{n-k+m} \left[n-k+m | \text{odd} \right] \right) \\ =\pi \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m=0 \\ n+2m=l}}^\infty \frac{c_n\left(b(b-1)\right)^m}{m!} + 2i \lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k,p=0 \\ k+m=l \\ n-k+m=2p+1}}^\infty \frac{c_n R^{n-k+m} \left(b(b-1)\right)^m}{(n-k+m)m!} \, .$$ Because of the limit, $p$ starts at $0$. Now equating the expressions found for the LHS and RHS and canceling common terms, it follows that each $\omega^l$-coefficient has to vanish identically $$0 = \lim_{R\rightarrow \infty} \sum_{\substack{n,m,p=0 \\ n+2m=l+2p+1}}^\infty \frac{c_n R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \\ = \lim_{R\rightarrow \infty} \sum_{p=0}^\infty \sum_{0\leq 2m \leq l+2p+1} \frac{c_{l+2p+1-2m} R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, . \tag{2}$$ If $b=0$ or $b=1$, the $m$-sum disappears except for $m=0$ and so $$\lim_{R\rightarrow \infty} \sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} = 0 \quad , \quad \forall l \in \mathbb{Z}_{\geq 0} \, . \tag{3}$$
Since the limit must exist for all $l\geq 0$, this is only possible if $$c_k =0 \quad , \quad \forall k \in \mathbb{N}$$
Note however, that in this case, there is no condition on $c_0$. Therefore $h(x)=\text{ const.}$, depending neither on $x$ nor on $\omega$.
In the case $b\neq 0$ and $b\neq 1$, the requirement (2) also forces $c_0=0$.
Some remarks to (3): If I set $$F_l(R)=\sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} \, ,$$ then (3) implies $$\lim_{R\rightarrow \infty} F_l'(R) = 0 \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, .$$ From this a relationship of $F_l'(R)$ to $F_0'(R)$ (for even $l=2m$) or $F_1'(R)$ (for odd $l=2m+1$) can be found $$F_{2m,2m+1}'(R) = \frac{F_{0,1}'(R)}{R^{2m}} - \frac{1}{R^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} R^{2k} \, .\tag{4}$$ Therefore, if a set of coefficients (not all zero) is found s.t. $F_{0,1}'(R)$ vanishes in the limit (e.g. $c_{2k+1,2k+2}=\frac{(-1)^k}{k!}$), then $F_l'(R)$ vanishes for all $l$. From this and (4) we can then derive a contradiction to (3), i.e. there exists at least one $l>1$ s.t. $$\lim_{R\rightarrow \infty} |F_l(R)|>0 \, .$$
Integrating (4) from $R$ to $\infty$ and assuming $F_{2m,2m+1}(R)$ vanishes in the limit $$0=\lim_{r \rightarrow \infty} F_{2m,2m+1}(r) = F_{2m,2m+1}(R) + \int_{R}^\infty F_{2m,2m+1}'(r) \, {\rm d}r \\ = F_{2m,2m+1}(R) + \int_{R}^\infty {\rm d}r \left( \frac{F_{0,1}'(r)}{r^{2m}} - \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right) \, ,$$ we see that for $R\rightarrow 0$, $F_{2m,2m+1}(R)$ goes to $0$, while the second term under the integral is only the polynomial approximation to $F_{0,1}'(R)$. The latter vanishes asymptotically, while the polynomial approximation does not. Hence, you will find $m$ s.t. $$\left|\int_{R}^\infty {\rm d}r \, \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right| \neq \left|\int_{R}^\infty {\rm d}r \, \frac{F_{0,1}'(r)}{r^{2m}} + O(R) \right|$$ for $R\rightarrow 0$.