[tex]u(t-a)=\begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}[/tex]
Then
[tex]L(u(t-a))= \int_0^\infty u(t-a)e^{-st} dt= \int_a^\infty e^{-st}dt[/tex]
Let v= t- a. Then t= v+ a, dv= dt, when t= a, v= 0, and when t= [itex]\infty[/itex], v= [itex]\infty[/itex]. The integral becomes
[tex]L(u(t-a)= \int_0^\infty e^{s(v+a)}dv= \int_0^\infty e^{sv}e^{sa}dv[/tex]
[tex]= e^{sa}\int_0^\infty e^{sv}dv= e^{sa}e^{-st}dt[/tex]
and that last integral is the Laplace transform of 1.
That u(t-a) step function multiplying a function basically multiplies the Laplace transform of the function by [itex]e^{as}[/itex].