Yes they are implying that. Applying the Laplace transform to a convolution gives a product. Write
$\mathcal{L}\{f*g\}(s):=\int_0^\infty \left(\int_0^x f(u)g(x-u)du\right)e^{-sx}dx=\iint_D f(u)g(x-u)e^{-sx}dudx$
The region of integration in the $xu$-plane is $D=\{(x,u):0\le u\le x\}$, an infinite triangle. Writing out the substitution $v=x-u$ our region of integration in the $uv$-plane is simply $(0,\infty)\times(0,\infty)$, so
$=\iint_{(0,\infty)^2} f(u)g(v)e^{-s(u+v)}dudv=\int_0^\infty f(u)e^{-su}du\int_0^\infty g(v)e^{-sv}dv=\mathcal{L}\{f\}(s)\cdot \mathcal{L}\{g\}(s).$
Note the Jacobian determinant of the transformation $(x,u)\mapsto(u,x-u)$ is simply $1$ (in abs. value).