I've noticed that it's often true that if: $$\mathcal{L}(f(t)\theta(t)) = F(s), \qquad Re(s)>0$$ then: $$\mathcal{L}(f(t)(\theta(t) - 1)) = F(s), \qquad Re(s)<0$$ For example it is true for $f(t)=sin(at)$
But is this a general rule?
If it where it would mean that if we have a table of Laplace transforms for functions defined on $(0,\infty)$ we could use the same table of one-sided transforms for functions defined on $(-\infty,0)$, as long as we multiply the function we want to transform by $-1$ first.