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 $$e^{-ct}\phi(t)\in L^2(0,\infty)\qquad\qquad\qquad(1) $$ for some $c\in \mathbb R$. In this case the Laplace transform $$f(s)=\int_{0}^{\infty}e^{-st}\phi(t)dt\qquad\qquad\qquad(2)$$ can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$. Moreover, it is easy to check that $$\sup\limits_{\sigma>c}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau<\infty.\qquad(3)$$ Now rewriting $(2)$ as $$f(\sigma+i\tau)=\int_{0}^{\infty}e^{-it\tau}e^{-\sigma\tau}\phi(t)dt,$$ we observe that $f$ is nothing more than just the Fourier transform of the function $e^{^{-\sigma t}}\phi(t)$ (trivially extended for by $0$ to $t\leq 0$) which belongs 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 holds
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}|f(\sigma+i\tau)|^2d\tau=\int_{0}^{\infty}e^{-2\sigma t}|\phi(t)|^2dt,$$ so there is a complete analogy with the standard Fourier transform.
Executive summary. A function $f$ is a the Laplace transform of some function $\phi$ satisfying condition (1), if and only if it can be extended analytically to the right half-plane $\{s=\sigma+i\tau:\ \sigma>c\}$ and condition (3) holds. This class of functions is known as the Hardy space$H^2(\sigma>c)$ on a (right) half-plane.
