The Coulomb law can be applied in the following manner by considering the continuous charge distribution to consists of infinitesimally small point charges and writing it as integral:
$$ \vec E(\vec r) = \sum_i q_i \frac{\vec r_i - \vec r}{\left|\vec r_i - \vec r\right|^3} = \sum_i \rho(\vec r_i) \Delta V \frac{\vec r_i - \vec r}{\left|\vec r_i - \vec r\right|^3} = \int d^3r' \rho(\vec r') \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3}. $$
Now, we can apply the Gauß integral theorem to prove:
\begin{align*}
\int_{\partial V} d\vec S \cdot \vec E &= \int_V d^3r\, \nabla \cdot \vec E(\vec r) = \int_V d^3r \nabla_{\vec r} \cdot \int d^3r' \rho(\vec r') \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} \\
&= \int_V d^3r \cdot \int d^3r' \rho(\vec r') \nabla_{\vec r} \cdot \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} = 4\pi \int_V d^3r\,\rho(\vec r)
\end{align*}
Here we use the formula $\nabla_{\vec r} \cdot \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} = 4\pi \delta(\vec r - \vec r')$. (You can easily prove that it is zero for $\vec r \ne \vec r'$ and then use the Gauß integral law to compute the strength of the $\delta$-peak by computing the flux.)
By replacing $\rho(\vec r) = q \delta(\vec r)$ we can consider a point charge and see that the Coulomb law implies the Gauß law.
The converse, however, is not true in general: The Gauß law holds in full electrodynamics, but there are additional electric fields induced by changing magnetic fields. You need to assume the electrostatic limit, where $\nabla \times \vec E = 0$ and that the field vanishes sufficiently fast at infinity to derive the Coulomb law from the Gauß law.
