Here is some hope. (For simplicity, when referring to the unit sphere, I really mean only the part of the unit sphere contained in the first quadrant $\mathbb{R}^n_+$.)
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}\cdot \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 $p\in P_s$. This 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. $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\vp}{\vec{p}}$ $\newcommand{\vx}{\vec{x}}$ $\newcommand{\vy}{\vec{y}}$
Remark 1 Let me denote by $Q_x$ the quadrant $\mathbb{R}^n_+$ in the $x$-space and by $Q_p$ the same quadrant but viewed in the $p$-space. Consider the kernel $K_\alpha:Q_p\times Q_x\to\bR$
$$ K_\alpha(\vec{p},\vec{x}) = e^{-(\vp\cdot {\vx})^\alpha}. $$
and its dual $K_\alpha^*:Q_y\times Q_p\to\bR$
$$ K_\alpha^*(\vec{y},\vp) = e^{-(\vec{y}\cdot\vp)^\alpha}. $$
The kernel $K_\alpha$ defines the generalized Laplace transform $L_\alpha$ when applied to functions on $Q_x$ that decay sufficiently fast. The convolution of the $K_\alpha^*$ with $K_\alpha$ is the kernel
$$ N_\alpha(\vy,\vx)=K_\alpha^* \ast K_\alpha(\vy,\vx)=\int_{Q_p} e^{-(\vy\cdot\vp)^\alpha-(\vp\cdot\vx)^\alpha} d\vp. $$
Note that
$$ N_\alpha(\vx,\vy)=N_\alpha(\vy,\vx)>0, \;\; \forall \vx,\vy\in Q$$
and if $f, g:Q\to\bR$ decay sufficiently fast at $\infty$ we have
$$\int_Q\int_Q N_\alpha(\vx,\vy) f(\vx)g(\vy) d\vx d\vy = (L_\alpha[f],L_\alpha[g])_{L^2(Q_p)}. $$
The integral operator $T_\alpha$ defined by $N_\alpha$ is symmetric and nonnegative, i.e.,
$$ (T_\alpha f,f)\geq 0,\;\; \forall f\in L^2(Q). $$
To check the injectivity of $L_\alpha$ one needs to show that
$$ (T_\alpha f,f)=0,\;\; f\in L^2(Q) \Rightarrow f=0. $$
Comment There seems to be a problemproblems with MathJax because I get distorted display and I cannot find my TeX errors. I'll post as is and Edit later.