$F(s)=\int f(t) e^{-s t}~dt$
$G(s)=\int f(2t) e^{-s t}~dt$
$u=2t, du=2dt,~t=u/2~,dt=du/2$
$G(s)=\frac 1 2 \int f(u) e^{-s u/2}~du=\frac 1 2 \in f(u) e^{-(s/2)u}~du=\frac 1 2 F(s/2)$
Uh, you know that "L(f)= 1/s" is only true for the specific function, f(t)= 1 for all t, don't you? Saying that "L{f(t)}= 1/s" is meaningless.
I hope, at least, that you know that .
So that . Now make the substitution u= 2x in that integral.