No, this cannot be true if $f$ is just $C^\infty$. Let $u=e^{(\Delta+f)t}u_0=e^{\Delta t}v_0$. Then, by well known properties of the heat equation, $u$ and $u_t$ are spatially analytic for $t>0$. But this implies that $f$ is also analytic everywhere where $u\neq 0$.

No, this cannot be true if $f$ is just $C^\infty$. Let $u=e^{(\Delta+f)t}u_0$. At $t=1$, $u=e^{\Delta+f}u_0=e^\Delta v_0$ for some $v_0$. Then, by well known properties of the heat equation, $u$ is spatially analytic. Moreover, $u_t=e^{\Delta+f}(\Delta+f)u_0$. If $u_0$ is sufficiently smooth, then $( \Delta+f)u_0$ is in $L^2$, so $u_t$ would have to equal $e^\Delta w_0$ for some $w_0$. This implies that both $u$ and $u_t$ are analytic, which makes $f$ analytic wherever $u\neq 0$.