The answer depends on the class of functions $\phi(t):(0,\infty)\to\mathbb R$ where you want to define the Laplace transform. A standard assumption is that
$$e^{-ct}\phi(t)\in L^2(0,\infty)\tag{1}\label{1} $$
for some $c\in \mathbb R$. In this case the Laplace transform
$$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\tag{2}\label{2}$$
can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that
$$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\tag{3}\label{3}$$
Now rewriting \eqref{2} as
$$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$
we observe that $f$ is just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended by $0$ to $t\leq 0$) belonging to $L^2(\mathbb R)$ for $\sigma=c$ and to $L^1(\mathbb R)\cap L^2(\mathbb R)$ for $\sigma>c$. Taking the
inverse Fourier transform, we get that
$$ e^{-\sigma t}\phi(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t>0,$$
and
$$0=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{it\tau}f(\sigma+i\tau)d\tau,\qquad t<0,$$
or, equivalently,
$$\lim\limits_{T\to\infty}\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}e^{st}f(s)ds=\begin{cases} \phi(t), & t>0 \\\ \\\ 0, & t\leq 0. \end{cases} $$

One can show also that the Parseval identity

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt$$
holds, so there is a complete analogy with the standard Fourier transform.

**Executive summary.** A function $f$ is the Laplace transform of some function $\phi$ satisfying condition \eqref{1}, if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition \eqref{3} holds. This class of functions is known as the Hardy space on a (right) half-plane.

assumethat the analytic function $f$ can be represented by some Laplace transform, then the Bromwich inversion integral does work under reasonable conditions. $\endgroup$