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 libear 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 is the complement great spheres
$$\{ \vec{p};\;\; |\vec{p}|=1,\;\;\vec{p}\cdot(\vec{x}_i-\vec{x}_j)=0 \},\;\;1\leq i <j\leq N. $$
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
$$\log f(t\vec{p}) \sim -t^\alpha(\vec{p}\cdot\vec{x}_{i(s)})^\alpha. $$
Hence
$$ (\vec{p}\cdot\vec{x}_{i(s)} )^\alpha =- \lim_{t\to\infty} \frac{\log f( t\vec{p})}{t^\alpha}. $$
This determines $\vec{p}\cdot\vec{x}_{i(s)}$ for any $p\in P_s$, and thus determines $\vec{x}_{i(s)}$. The constant $c=c_{i(s)}$ 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.