Let $\mu$ be a probability measure on $\mathbb R$ and $$\mathcal L_\mu(t):=\int e^{-tx}\:\mu({\rm d}x)\;\;\;\text{for }t\in\mathbb R$$ denote the Laplace transform of $\mu$. Assume $$\mu((-\infty,0))=0\tag1$$ and $$-\ln\mathcal L_\mu(t)=\alpha t+\int 1-e^{-tx}\:\nu({\rm d}x)\;\;\;\text{for all }t\ge0\tag2$$ for some $\alpha\ge0$ and some $\sigma$-finite measure $\nu$ on $[0,\infty)$ with $$\nu(\{0\})=0.\tag3$$
I want to show that $$\alpha=\sup\left\{x\ge0:\mu([0,x))=0\right\}\tag4.$$
Intuitively, the claim shouldn't be hard to prove. And I think it's related to the support $\operatorname{supp}\mu$ of $\mu$. In the present context, this notion should reduce to $$\operatorname{supp}\mu=\left\{x\ge0:\mu((x-\varepsilon,x+\varepsilon))>0\text{ for all }\varepsilon>0\right\}\tag5.$$ Maybe we can show that the right-hand side of $(4)$ is equal to $\inf\operatorname{supp}$. However, even if it's possible to show that, I still wouldn't know how we can conclude $(4)$ from this result.
Define
$$\beta := \sup\{x \geq 0\::\: \mu([0,x))=0\},$$
then
$$\mathcal{L}_{\mu}(t) = \int_{[\beta,\infty)} e^{-tx} \, \mu(dx) = e^{-\beta t} \underbrace{\int_{[0,\infty)} e^{-tx} \, \tilde{\mu}(dx)}_{\mathcal{L}_{\tilde{\mu}}(t)} \tag{1}$$
for the shifted measure $\tilde{\mu}$ defined by $\tilde{\mu}(B) := \int 1_{B}(x-\beta)\, \mu(dx)=\mu(B+\beta)$. Note that, by construction, $\tilde{\mu}([0,x])>0$ for any $x>0$. Since $\tilde{\mu}$ is also a probability measure with $\tilde{\mu}(-\infty,0)=0$, its Laplace transform satisfies
$$- \ln \mathcal{L}_{\tilde{\mu}}(t) = \tilde{\alpha} t + \int (1-e^{-tx}) \tilde{\nu}(dx) \tag{2}$$ for some $\tilde{\alpha} \geq 0$ and some $\sigma$-finite measure $\tilde{\nu}$ on $[0,\infty)$. By $(1)$, we are done if we can show that $\tilde{\alpha}=0$. As $1-e^{-tx} \geq 0$, it follows from $(2)$ that
$$\mathcal{L}_{\tilde{\mu}}(t) \leq e^{-\tilde{\alpha} t}.$$
On the other hand, the monotonicity of the exponential function yields
$$\mathcal{L}_{\tilde{\mu}}(t) \geq \int_{[0,r]} \underbrace{e^{-tx}}_{\geq e^{-rt}} \, \tilde{\mu}(dx) \geq e^{-rt} \tilde{\mu}[0,r],$$
implying
$$\tilde{\mu}[0,r] e^{-rt} \leq e^{-\tilde{\alpha} t}$$ for all $t \geq 0$. If $\tilde{\mu}([0,r])>0$, then this gives $\tilde{\alpha} \leq r$. Recalling that $\tilde{\mu}([0,r])>0$ holds, by construction, for any $r>0$, we conclude that $\tilde{\alpha}=0$.