Let $K(x)$ denote the Schwartz kernel of the Fourier multiplier $\nabla e^{-t(-\Delta)^\alpha}$. It's straightforward to check from Fourier inversion and Plancherel's theorem that for $t>0$,
$$K(x) = (2\pi)^{-3}\int_{\mathbb{R}^3} i\xi e^{-t|\xi|^{2\alpha}}e^{i\xi\cdot x}d\xi,\qquad\forall x\in\mathbb{R}^3.$$
I claim that $K\in L^1(\mathbb{R}^3)$. To see this, observe that for any multi-index $\beta=(\beta_1,\beta_2,\beta_3)$ of order $|\beta|=4$ and $x\in\mathbb{R}^3$ such that $x^\beta = x_1^{\beta_1}x_2^{\beta_2}x_3^{\beta_3}\neq0$, we have that
$$K(x)=\frac{1}{x^\beta (2\pi)^3}\int_{\mathbb{R}^3}i\xi e^{-t|\xi|^{2\alpha}} (\partial_{\xi}^\beta e^{i\xi\cdot x})d\xi.$$
Now integrate by parts in $\xi$. The most singular term in the application of the Leibnitz rule comes from repeatedly differentiating $|\xi|^{2\alpha}$. You can show that
$$|K(x)|\lesssim_\alpha \frac{1}{|x^\beta|}\int_{\mathbb{R}^3}|\xi|^{2\alpha-3}e^{-t|\xi|^{2\alpha}}d\xi\lesssim \frac{t^{-1/2\alpha}}{|x^\beta|},$$
where the last expression follows from dilation invariance of Lebesgue measure. Since $|\beta|>3$, the desired conclusion follows.

Regarding negative-order Besov spaces, try playing around with the fact that Littlewood-Paley projectors reduce Fourier multipliers to multiplication by a function of a dyadic frequency.