Suppose we have two functions of $t$, $f(t)$ and $g(t)$.
Letting $\mathcal{L}\{f(t)\} = F(s)$ and $\mathcal{L}\{g(t)\} = G(s)$, I know that:
$$\mathcal{L}\{f(t) \star g(t)\} = F(s) \cdot G(s),$$
but then I thought naturally it should be true that, applying the inverse Laplace transform to both sides, we get:
$$f(t) \star g(t) = \mathcal{L}^{-1}\{ F(s) \cdot G(s) \}.$$
However, this appears to be not the case, or at least for the question:
$$\mathcal{L}^{-1}\left\{ \frac{1}{s^2} \cdot \frac{1}{s-3} \right\},$$
where I would have expected the answer to be:
$$\mathcal{L}^{-1} \left\{ \frac{1}{s^2} \cdot \frac{1}{s-3} \right\} = t \star e^{3t}.$$
However this is not the case. I do not understand why, however. Is my issue that I am assuming $\mathcal{L}^{-1}\{\mathcal{L}\{h(t)\}\} = h(t)$? Or is it deeper than that?
We have $$ t*e^{3t}=\int_0^t (t-\tau)e^{3\tau}\,d\tau=\frac{1}{9}(e^{3t}-1-3t) $$ and that seems to be the same one gets doing partial fraction decomposition on $1/(s^2(s-3))$, $$ \frac{1}{s^2(s-3)}=\frac{1}{9}\Bigl(\frac{1}{s-3}-\frac{3}{s^2}-\frac{1}{s}\Bigr), $$ and then inverse Laplace transform, yielding $\frac{1}{9}(e^{3t}-1-3t)$ again.
So, everything seems to be OK.