The property says: $$L[f(t-a)] = e^{-as} * F(s)$$
Standard proof goes as:
$$L[f(t-a)] = \int_{0}^{\infty}f(t-a)*e^{-st}dt$$
Now, we make a change of variables, assume $u=t-a$ then $t=u+a$ and $du=dt$
So, now the integral is $$\int_{0}^{\infty}f(u)*e^{-s(u+a)}du $$
Integral limits from $u = 0-a = -a$ to $t = \infty-0 = \infty$
However, the function $f(t-a) = 0$ for $t \lt a$ and so the boundaries of the integral goes from $0$ to $\infty$
The integral reduces to $\int_{0}^{\infty}f(u)*e^{-su}*e^{-sa}du = e^{-as} ∫_{0}^{\infty}f(u)*e^{-su}du$
The book says therefore, $L[f(t-a)] = e^{-as} F(s) $
The book states that $\int_{0}^{\infty}f(u)*e^{-su}du$ is, by definition, the Laplace transform and I understand this. My question is:
Shouldn't this $F(s)$ be the Laplace transform of $f(u)$ and not $f(t)$? It says it's the Laplace transform of $f(t)$ and this is the part I can't understand
Your notation is a bit difficult for me to get the point. I am posting the version which I teach at the class. Let $h(x)=e^{cx}f(x)$ ($c$ is $-a$ in yours above) then for all $s>a+c$ we have $$H(s)=\int_0^{\infty}e^{-sx}e^{cx}f(x)dx=\int_0^{\infty}e^{-(s-c)x}f(x)dx=F(s-c)$$ in which $f(x)=\mathscr{L}^{-1}(F(s)).$