You have to be clear about which flavor of the Laplace transform you're talking about. If you consider the bilateral Laplace transform
$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt\tag{1}$$
then the relationship
$$\mathcal{L}\{f(t+a)\}=e^{as}F(s)\tag{2}$$
clearly holds, also for $a>0$.
However, if you consider the unilateral Laplace transform
$$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt\tag{3}$$
then for $a>0$
$$\mathcal{L}\{f(t+a)\}=e^{as}\int_a^{\infty}f(t)e^{-st}dt\tag{4}$$
which is generally not equal to $e^{as}F(s)$ (unless $f(t)=0$ in $[0,a]$).
In your example $f(t)=t^2$ you implicitly use the unilateral Laplace transform, so $(2)$ doesn't hold.