Laplace Transform of $f(t+a), a>0$ where $f(t)$ is not periodic
Let $s = \sigma + j\omega$, the inverse Laplace transform of $f(t+a)$ is given by
$$f(t+a) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{s(t+a)} \mathrm{d}s = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{sa}e^{st} \mathrm{d}s.$$

Hence the bilateral Laplace transform of $f(t+a)$ is $F(s)e^{sa}$ where $F(s)$ is the Laplace transform of $f(t)$. For the unilateral case, see Matt L. answer.



This is sometimes called the shifting theorem. See Theorem 12.16 here.