Sum of Convolutions, Laplace Transform?

Question

Suppose that we have a function $g(t)$ that can be computed from a function $h(t)$ from

$g(t) = h(t) + h(t)*h(t) + h(t)*h(t)*h(t) + ...$

where $*$ is convolution. We have the following conditions:

1. Both functions are defined over $t \geq 0$

2. $g(t), h(t) < 1$ for all $t$.

3. $\int_{0}^{\infty} h(t) dt = 1$, i.e. $h(t)$ is a probability density function.

Is it possible to compute $h(t)$ in terms of $g(t)$? So far, I could think of taking a Laplace transform from both sides. Therefore

$G(s) = H(s) + H(s)^2 + H(s)^3 + ... $

If $G(s)$ is finite (for example by taking the Laplace over finite amount of time?), then we have $|H(s)|<1$. We can write

$G(s) \approx \frac{H(s)}{1-H(s)}$

Therefore

$H(s) \approx \frac{G(s)}{1+G(s)}$

$h(t) \approx \mathcal{L}^{-1}[\frac{G(s)}{1+G(s)}]$

where $\mathcal{L}^{-1}$ is Laplace inverse. Is this approach correct? Can we compute $h(t)$ in terms of $g(t)$ directly? Does $h(t)$ somehow relate to $\exp (g(t))$?