$$\require{cancel}$$
$\textbf{L}\left(e^{-at} y(t) \right) = \bar{y}(s+a).$
For example, from the table, we have $$\textbf{L}(t) = 1/s^2$$. The shifting theorem tells us that $$\textbf{L}\left(te^{-at} \right) = 1/(s+a)^2$$. I'm sure you will now want to expand your table even more. Or you may want to go the other way, and cut down the table a bit! After all, you know that $$\textbf{L}(1) = 1/s$$. The shifting theorem, then, tells you that $$\textbf{L}(e^{at}) = 1/(s-a)$$, so that entry in the table is superfluous! Note that you can use the theorem to deduce either direct or inverse transforms.