The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law.
More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law
$$
\nabla\cdot\mathbf{E}=\rho/\epsilon_0\quad\Leftrightarrow\quad \oint_S\mathbf{E}\cdot\mathrm{d}\mathbf{a}=Q/\epsilon_0,
$$
where $Q$ is the total charge enclosed by the (arbitrary) surface $S$.
To derive Coulomb's law, consider the electric field of a single point particle, with nothing else in the universe. Because of isotropy (which must be added as an additional postulate), the electric field at a sphere of radius $r$ centred on the charge must be radial and with the same magnitude throughout. That means the integral is trivial and the electric field must be
$$\mathbf{E}=\frac{Q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}.$$
Coupled with Lorentz's force law at zero velocity for the test particle (since Coulomb's law only holds in electrostatics) this yields Coulomb's law.
It is not obvious that this highly symmetric situation can give the general electrostatic force for multiple particles. This follows from the superposition principle, which is very much at the heart of classical electrodynamics, and which can be obtained from the linearity of Maxwell's equations. This gives you the field for a single source; add the fields for all the individual sources and you'll get the field for the collection of sources.
