Do Inverse Laplace transforms satisfy the Convolution Theorem too? The Laplace transform of a function $f$ is defined as
$$
F(s) \ = \ \mathcal{L}[f](s) \ = \ \int_0^\infty dt \ f(t) e^{-st}
$$
Whereas we can write the inverse Laplace transform in terms of the Bromwich integral
$$
f(t) \ = \ \mathcal{L}^{-1}[F](t) \ = \ \frac{1}{2\pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} dt \ F(s) e^{+st}
$$
where $\gamma$ is a number which is greater than the real part of all the singularities of $F(s)$, and such that $F(s)$ is bounded along the above line.
The convolution theorem states that for functions $f(t)$ and $g(t)$ (where $F(s):=\mathcal{L}[f](s)$ and $G(s):=\mathcal{L}[g](s)$) we have
$$
\mathcal{L}[f \ast g](s) \ = \ F(s) G(s)
$$
where $(f \ast g)(t):=\int_0^t dt'\ f(t')g(t-t')$.
My Question: 
Does a similar theorem hold for the inverse Laplace transform? Meaning can we say 
$$
\mathcal{L}^{-1}[F \ast G](t) \ = \ f(t) g(t)
$$
where $(F \ast G)(s)=\int_0^s ds'\ F(s')G(s-s')$.
 A: No, not because the convolution theorem doesn't hold, but because the convolution of $F,G$ isn't defined like you did. The correct one is $$F \ast G(s) = \frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(z) G(s-z) dz$$ for any $r > 0, \Re(s) > r$ which is well-defined for example when assuming $f,g$ are bounded so their Laplace transform is $L^2$ on vertical lines.
Assuming $f,g$ are $C^1$ and they vanish at $t=0$ then $F,G$ are $L^1$ on vertical lines and for $c > r$ using absolute convergence / Fubini to change the order of integration $$\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty} F \ast G(s)e^{st}ds = \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty}\frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(s-z) G(z)e^{st}dz ds $$
$$= \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty}\frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(s-z) e^{(s-z)t}G(z)e^{zt}dz ds $$
$$= \frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty} F(s-z) e^{(s-z)t}G(z)e^{zt}ds dz$$ 
$$= \frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty}\frac{1}{2i\pi}\int_{c-r-i\infty}^{c-r+i\infty} F(u) e^{ut}G(z)e^{zt}du dz$$
$$= (\frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty} F(u) e^{ut}du )(\frac{1}{2i\pi}\int_{c-r-i\infty}^{c-r+i\infty}G(z)e^{zt} dz)$$
$$ = f(t)g(t)$$ 
