Short answer:
The short answer is that the transition from mixed salad dressing to two layers is exothermic, and this heat release creates an entropy increase. For all intents and purposes, the process is analogous to an exothermic chemical reaction, like combustion.
Similarly, a container filled with ball bearings can spontaneously settle into a hexagonal close-packing arrangement. Like the salad-dressing case, this gives the visual appearance of greater order, but in fact it is not a violation of the law $\Delta S_\text{univ}\geq 0$, as energy is released upon settling, which is converted into disordered heat.
Slightly longer answer:
I'm terrible at thermodynamics so there may be several corrections needed to make the following rigorous, but you can try to make things explicit as follows: let the salad dressing be contained in a rigid, thermally-conducting container under the influence of gravity. The total energy of the system can be written as
$$U=m_\text{w}\overline{U}_\text{w,bulk}+m_\text{o}\overline{U}_\text{o,bulk}+\int_V\rho(\mathbf{r})V(\mathbf{r})\,d\mathbf{r}+\int_{S_d}\gamma_\text{w,o}\,dS+\int_{S_c}\gamma(S)\,dS$$
where $\overline{U}_\text{w,bulk}$ is the total bulk internal energy per mass of the water (and similar for $U_\text{o,bulk})$, $m_\text{w}$ and $m_\text{o}$ are the total water and oil masses, $\rho(\mathbf{r})$ is the liquid density at location $\mathbf{r}$ in the container, $V(\mathbf{r})$ is the gravitational potential, $S_d$ is the set of oil-water interfaces, $\gamma_\text{w,o}$ is the oil-water surface tension, $S_c$ is the boundary of the container walls, and $\gamma(S)$ is the liquid-wall surface tension of the type of liquid at boundary location $S$.
In essence, the first and second terms describe the bulk (volumetric) energy of the water and oil, the third considers the gravitational energy of the system, the fourth considers the energy due to the oil-water interfaces, and the fifth considers the energy due to the liquid-container interface.
Linearizing gravity as $V(\mathbf{r})\approx g|\mathbf{r}|$, $U$ can be rewritten as
$$U=m_\text{w}\overline{U}_\text{w,bulk}+m_\text{o}\overline{U}_\text{o,bulk}+\rho_\text{w}m_\text{w}\langle h_\text{w}\rangle+\rho_\text{o}m_\text{o}\langle h_\text{o}\rangle\\+\gamma_\text{w,o}A_d+\gamma_\text{w,c}A_{\text{w,c}}+\gamma_\text{o,c}A_{\text{o,c}}$$
where $\langle h_\text{w}\rangle$ and $\langle h_\text{o}\rangle$ are the expectation values of the water and oil height inside the container, $A_d$ is the total oil-water droplet interface area, and $A_{\text{o,c}}$ and $A_{\text{w,c}}$ are the total areas of contact the oil and water make with the container walls and $\gamma_\text{w,c}$ and $\gamma_\text{o,c}$ are the water-container and oil-container surface tensions.
A change of configuration creates a change in internal energy
$$\Delta U=\rho_\text{w}m_\text{w}\langle \Delta h_\text{w}\rangle+\rho_\text{o}m_\text{o}\langle \Delta h_\text{o}\rangle+\gamma_\text{w,o}\Delta A_d+\gamma_\text{w,c}\Delta A_{\text{w,c}}+\gamma_\text{o,c}\Delta A_{\text{o,c}}.$$
Note that $\Delta U<0$ for a transition from mixed salad dressing into two separated layers.
The excess energy is converted into heat, ie $\Delta U=\Delta q$, which is then lost to the surroundings through the container walls. The entropy change is then
$$\Delta S=\Delta S_\text{sys}+\Delta S_\text{surr}=-\frac{\Delta U}{T_\text{sys}}+\frac{\Delta U}{T_\text{surr}}>0$$ since $T_\text{sys}>T_\text{surr}$ for heat transfer to occur.
(Apologies if I botched up the fundamental thermodynamic relations in the previous paragraph).