Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only
There are many examples of such reactions, for instance:

$$
\begin{align}
\ce{S (s) + O2 (g) &amp;&lt;=&gt; SO2 (g)}\\
\ce{C (s) + 2 H2 (g) &amp;&lt;=&gt; CH4 (g)}\\
\ce{Ni (s) + 4 CO (g) &amp;&lt;=&gt; Ni(CO)4 (g)}
\end{align}
$$

However, your assumption 


  all that is considered is which side the solid/liquid is on, not the moles of gas present on either side


is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.