Degenerated common source output resistance by inspection
My question is answered. I wrote up some notes fully explaining what's going on with this inspection analysis and clearing up some other details that were confusing me. Hopefully others will find them helpful.
The short answer is that when you use visual shortcuts like drawing the \$r_o\$ in parallel with the MOSFET symbol, you're separating out the MOSFET into two parts: one part which holds the dependence on \$v_{ds}\$ only (\$r_o\$) and one part that holds the dependence on \$v_{gs}\$ only (the traditional MOSFET symbol). At the source, the resistor \$ R_S \$ senses \$ \Delta V_{RS} \$ and carries away a current proportional to it (and no other voltages). The MOSFET symbol (since it &quot;holds&quot; only the VCCSs \$g_m v_{gs} \$ and \$ g_m v_{bs}\$) will also sense the voltage \$ \Delta V_{RS}\$ and carry away from the source a current  proportional to it (and no other voltages). This means the MOSFET symbol has the exact same function as the resistor \$R_S\$, so you can model the MOSFET symbol as a resistor and put it in parallel as shown.