Please find the flaw in my reasoning.
Given a function $f(t)$, the Laplace transform is $F(s) = \int_0^\infty dt f(t) e^{-st}$. And the inverse is $f(t) = \frac{1}{2\pi i} \int_\gamma ds F(s) e^{st}$ where $\gamma$ indicates a path in the region of convergence.
Define a time shifted version $g(t) = f(t-a) u(t-a)$, where $u$ is the step function, and $a>0$.
You can prove that $G(s) = e^{-as}F(s)$ using the time shift property. But when I try to compute the inverse the step function is gone.
$g(t) = \mathcal{L}^{-1}(G(s))=\frac{1}{2\pi i}\int_\gamma ds G(s) e^{st}=\frac{1}{2\pi i}\int_\gamma ds F(s) e^{s(t-a)} = f(t-a)$
Where in the last step I used the definition of the inverse Laplace transform at the start of my question. So where did the step function go?
Edit: Added the assumption $a>0$.
You don't recover the Heaviside function inside $g(t)$ after inverting the Fourier transform, because you calculated $G(s)$ as if it was absent from $g(t)$ at the start. Indeed, one has : $$ G(s) = \int_0^\infty f(t-a)u(t-a)e^{-st}\,\mathrm{d}t = e^{-sa}\int_\color{red}{-a}^\infty f(t)u(t)e^{-st}\,\mathrm{d}t \neq e^{-sa}F(s) \mathrm{\;\;if\;\;} a<0 $$ Note that products of functions are mapped to convolution products by Fourier/Laplace transforms (although more caution has to be shown with the path of integration in the case of Laplace), i.e. $\mathscr{L}[f_1f_2] = F_1*F_2$.