Here is some hope. If $\mu$ is the Dirac  delta  concentrated at a point $\vec{x}$ then  $L[\mu](\vec{p})= e^{-(\vec{p}\cdot\vec{x})^\alpha}$, which shows that $\mu$ is determined by this  transform. More generally suppose $\mu$ is a linear combination of $\delta$-s

$$\mu=\sum_{k=1}^N c_k \delta(\vec{x}-\vec{x}_k). $$

In this case  we have 

$$L[\mu](\vec{p})=\sum_{k=1}^N c_k e^{-(\vec{p}\cdot\vec{x}_k)^\alpha}=:f(\vec{p}). $$

Consider the set $P$ of $\vec{p}$-s of length $1$ so that the numbers $\vec{p}\cdot\vec{x}_i$  are pairwise    disjoint.  $P$ is open and dense in the  unit sphere because it's the complement    in the unit sphere is the union of  the great spheres $\vec{p}\vec{x}_i=\vec{p}\cdot\vec{x}_j $, $i\neq j$.



     

The set  $P$ is made up of several connected components, chambers, $P_1,\dotsc, P_\nu$.    For  any chamber $P_s$, there exists $i(s)=1,\dotsc, N$ such that

$$\vec{p}\cdot \vec{x}_{i(s)} < \vec{p}\cdot \vec{x}_j,\;\;\forall j\neq i(s),\;\;\forall \vec{p}\in P_s. $$


Then as $t\to \infty$ we have

$$\frac{\log f(t\vec{p})}{t^\alpha} \sim -(\vec{p}\cdot\vec{x}_{i(s)})^\alpha. $$



This  determines $\vec{p}\cdot\vec{x}_{i(s)}$ for any $p\in P_s$, and thus determines the point $\vec{x}_{i(s)}$.     The constant $c_{i(s)}$ is then determined from the equality


$$c_{i(s)}= \lim_{t\to\infty} e^{t^\alpha(\vec{p}\cdot\vec{x}_{i(s)})^\alpha} f(t\vec{p}),\;\; \vec{p}\in P_s.$$

The function 

$$ g(\vec{p})= f(\vec{p})-c_s e^{-(\vec{p}\cdot \vec{x}_{i(s)})^\alpha} $$

is the generalized Laplace transform   of a  linear combination of $\delta$-s concentrated at $(N-1)$ points. 

Thus, the generalized Laplace transform  is  injective when restricted to the vector space spanned  by $\delta$-s.

**Comment** There seems to be a problem with MathJax   because I get    distorted display and I cannot find my TeX errors. I'll post as is and Edit later.