Yes, because $f\star g$ is indeed real-analytic (provided $g$ doesn't grow too fast at $\infty$), so that, being $0$ on an interval, it is $0$ everywhere. Then the Fourier transform of $f\star g$, which is a Gaussian times the Fourier transform of $g'$, is $0$, and as the Gaussian is nonzero everywhere this implies $g'=0$ as a distribution.

Yes, because $f*g$ is indeed real-analytic (provided $g$ doesn't grow too fast at $\infty$), so that, being $0$ on an interval, it is $0$ everywhere. Then the Fourier transform of $f*g$, which is a Gaussian times the Fourier transform of $g'$, is $0$, and as the Gaussian is nonzero everywhere this implies $g'=0$ as a distribution.