I'm looking for a proof that for a renewal process $\{N_t\}_{t\geq0}$ it is equivalent to have a linear renewal function and to be a Poisson process. Since a Poisson process is clearly a renewal process, I'm interested in the other direction.
Let $(X_i)_{i\in\mathbb N}$ be a sequence of i.i.d. random variables that are positive with probability $1$ and have finite expected value, the renewal sequence. Let $S_n = \sum_{i\in\mathbb N}^n X_i$, the time of the nth renewal. The stochastic process defined by $N_t=\sup\{n\in\mathbb N \vert S_n \leq t\}$ is called a renewal process.
The renewal function of $\{N_t\}$ is defined by $M(t) = \mathbb E[N_t]$.
Since a renewal process starts at $t=0$, I want to show that:
Let $\{N_t\}$ be a renewal process with renewal function $M$ and $\alpha>0$ such that $M(t)=\alpha t$. Then $\{N_t\}$ is a Poisson process with intensity $\alpha$.
Approach
Thanks to the comment, I learned that this can be solved by applying Laplace-Stieltjes transformations. I'm stuck in asserting by means of the renewal equation $$ \mu(t) = H(t) + \int_0^t \mu(t-x) \mathrm d F(x)\, , \quad t \geq 0 $$ for a uniformly bounded function $H$ and a distribution function $F$ (of the arrival times) that \begin{equation} \tag{1} \mu^*(t) = \frac{H^*(t)}{1-F^*(t)} \, , \quad t \neq 0 \end{equation} where $$ \mu^*(t) = \int_{-\infty}^{\infty} \mathrm e^{-tx} \mathrm d \mu(x) $$ is the Laplace-Stieltjes transform of $\mu$.
How do I arrive at $(1)$? If I'm not mistaken, we should have some form of linearity such that $$ \mu^*(t) = H^*(t) + \int_{-\infty}^{\infty} \mathrm e^{-tx} \mathrm d \left( \int_0^t \mu(t-x) \mathrm d F(x) \right) \, , $$ but I have trouble evaluating the last term. To come up with the desired result, it should evaluate to $\mu^*(t) F^*(t)$. That is assuming $(H+F)^*=H^*+F^*$ which I didn't find any reference to, but it seems immediately from the definition.