Consider the inverse Laplace transform of the following function \begin{align} e^{-s\tau}P(s) \end{align} from the $s$ domain to the $t$ domain. We know that the inverse Laplace transform of $e^{-s\tau}$ is $\delta(t-\tau)$; and the inverse Laplace transform of $P(s)$ is defined to be $p(t)$. Using the convolution theorem \begin{align} \mathcal{L}^{-1}[F(s)G(s)]=\int_0^\infty f(t-t')g(t')dt' \end{align} we write \begin{align} \mathcal{L}^{-1}[e^{-s\tau}P(s)]=\int_0^\infty\delta(t-t'-\tau)p(t')dt'=p(t-\tau)\theta(t-\tau) \end{align} We obtain the step function $\theta(t-\tau)$ by observing that since $t'\geq 0$ as set by the integration limit, and $t'=t-\tau$ as set by the delta function, we only have non-vanishing result when $t\geq \tau$. This expression arises when I was working with an emitter and a spatially separated receiver, which receives the signal after time $\tau$ has elapsed. It seems that the step function makes sense in this context: the term is "turned on" at $t=\tau$.
What I don't understand is this: since multiplication is commutative in the Laplace domain, I could just as easily write down the following convolution: \begin{align} \mathcal{L}^{-1}[P(s)e^{-s\tau}]=\int_0^\infty\delta(t'-\tau)p(t-t')dt'=p(t-\tau)\theta(\tau) \end{align} The argument of the step function changes because of a line of thought similar to the one above: $t'>0$ as set by the integration limit, $t'=\tau$ as set by the delta function, so we only have non-vanishing result when $\tau>0$.
Clearly the first and the second convolutions are different, and I'm guessing that the first one is correct, so what did I miss in the second one?