confused about time shifting property of Laplace Transform
If you have written the function correctly then its Laplace transform could be found very similary to your first example:

Given
$$x(t) = e^{-2 t} u(t-1)$$ 
its Laplace transform could be found as follows. First denote the signal

$$x_0(t) = e^{-2} e^{-2t} u(t) $$ 

then its obvious that 
$$x(t) =x_0(t-1) $$

Using the tables and properties to conclude:

$$X(s) = e^{-s} X_0(s) $$
$$X(s) = e^{-s} \frac{ e^{-2} }{s + 2} = \frac{ e^{-(s+2)} }{s + 2}   $$