i'm not sure how the title of your post is related to the content.
by definition, on the left side you have,
Y(s) = \mathcal{L} \left\{y(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} y(t) \,dt
and, also by definition,
X(s) = \mathcal{L} \left\{x(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt
and, it turns out that for LTI systems, after you show that the convolution operator and some impulse response is what relates the input x(t) to the output y(t), then
Y(s) = H(s) X(s)
where
H(s) = \mathcal{L} \left\{h(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} h(t) \,dt
and h(t) is the impulse response of the LTI system and completely characterizes the input/output relationship of the LTI system.
now, in your specific case,
Y(s) = \mathcal{L} \left\{y(t)\right\} = \mathcal{L} \left\{x(t-3)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t-3) \,dt
which, after you do a trivial substitution of variable of integration is
Y(s) = \int_{-\infty}^{+\infty} e^{-s(t+3)} x(t) \,dt = e^{-s \cdot 3} \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt = e^{-s \cdot 3} \cdot X(s)
which means that
H(s) = e^{-s \cdot 3}.
