Electronic – A switch can represent the unit step function

it also multiplies by the unit step function for reasons I fail to comprehend...

The problem asks that you give the solution for \$t \gt 0\$ but that still doesn't quite explain the presence of \$u(t)\$ in the solution. I think you would be fine if you don't have the \$u(t)\$ term as long as you indicate the solution is for \$t \gt 0\$.

The solution is using Kirchoff's Current Law to find ... something?

To find the voltage at node 1, the common connection of the three resistors. Knowing \$v_1\$ allows you to find the current through the capacitor via the series resistor thus

$$i_C = \frac{v_1 - v_O}{R}$$

You now have the voltage across and the current through the capacitor so, using the capacitor equation

$$C \frac{dv_O}{dt} = i_C = \frac{v_1 - v_0}{R}$$

you can write the differential equation to solve for \$v_O(t)\$

Now, you can go directly to the solution for a 1st order RC circuit if you know the time constant and, for that, you must find the equivalent resistance of the circuit connected to the capacitor.

The twist here is that you have a dependent source to deal with - the output of the opamp - so you must be careful in setting up the calculation for the equivalent resistance.

It's straightforward to show that the equivalent resistance is \$3R\$

I think the simulator is wrong. I would expect the capacitor to be charged to about 4.3 volts, allowing for an 0.7 Volt base to emitter drop in the transistor. (Although the transistor may fail due to excessive base current when the switch is first closed.)