The laplace transform for ramp function with a time delay can be expressed as follows

Step 1
Let the time delay is T,
So the ramp function will be
$f(t)=(t-T)u(t-T)$
Where
$u(t-T)=\{\begin{array}{ll}1& t>T\\ 0& t\end{array}$
Step 2
Let the Laplace transform of f(t) is F(s)
So the Laplace transform is
$F(s)={\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$
$\Rightarrow F(s)={\int }_T^{\mathrm{\infty }}T(t-T)e^{-st}dt$
let
x=t-T
$\Rightarrow dx=dt$
So we have
$F(s)={\int }_0^{\mathrm{\infty }}x{e}^{-s(x+T)}dx$
$\Rightarrow F(s)=e^{-sT}{\int }_0^{\mathrm{\infty }}x{e}^{-sx}dx$
$\Rightarrow F(s)=\frac{e^{-sT}}{s^2}$