Suppose $f \in C^{\infty}_c((-1,1))$ and assume that there exists constants $a,b>0$ such that $$ \bigg|\int_{\mathbb R} f(t) \,e^{\tau t^2+i\tau t}\,dt\bigg| \leq a\,e^{-b|\tau|},$$ for all $\tau \in \mathbb R$.
Does it follow that $f=0$?
The expression inside absolute value sign in the LHS is an entire function of $\tau$ (since it can be differentiated under the integral sign and derivative exists for all complex $\tau$). Your inequality shows that it tends to zero as $\tau\to\infty$ therefore it is zero by Liouville's theorem. Then Fourier inversion says that $f=0$.
To address some comments. The function is entire of exponential type. Since the indicator is negative for $\arg\tau\in(0,\pi)$ is must be $0$ by Phragmen-Lindelof (for all complex $\tau$).
Edit: I'm undeleting this now, though it may not be of much interest any more. I thought the argument was incomplete, but actually the missing step is quite easy, given a fact about density of polynomials on compact subsets of $\mathbb C$ (Mergelyan's theorem).
Thank you to fedja for pointing this out; see his comments to the linked question.
I think I now know how to provide all the details in Alexandre's strategy.
Consider $F(z)=\int f(t)e^{zt^2+izt}\, dt$. This is an entire function of exponential type. Its indicator function $h(\alpha)=\limsup_{r\to\infty} (1/r)\log |F(re^{i\alpha})|$ satisfies $h(0), h(\pi)\le -b$, by assumption. This is impossible for an $F\not\equiv 0$ since always $h(\alpha)+h(\alpha+\pi)\ge 0$ (see, for example, Levin's book for this).
So $F\equiv 0$ and also $$ F^{(n)}(0) = \int_{-1}^1 f(t) t^n(t+i)^n\, dt = 0 $$ for all $n\ge 0$. Since $\gamma(t)=t(t+i)$ is injective on $[-1,1]$, this implies that $f=0$.