RC time constant in parallel circuit 
$R_1 = 50k \Omega $
$ R_2 = 100k \Omega $
$ \mathcal{E} = 10V $
$ C = 10 \mu F $
So I want to find the $\tau$ time constant of this circuit after the switch is closed. 
The only issue is that when the switch is closed, the circuit becomes parallel, which means that simply finding $R_{eff}$ isn't as simple as doing $\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}$ then doing $R_{eff} C = \tau $.
From the answer in the textbook, it appears that the time constant $\tau$ for the circuit after the switch is flipped is just $R_2 C$ $ (100k \Omega \times 10 \mu F) $, which means that either there's a way to take $50k \Omega$ and $ 100k \Omega $ and turn it into $ R_{eff} = 100k \Omega $, or you ignore $R_1$.
I'm guessing that it has something to do with how the current changes as time progresses, since the capacitor will slowly allow less and less current to "pass" through as it charges.
Why does the time constant for the circuit not depend on $R_1$ when the switch is closed?
 A: When the switch is closed, we can easily see that the sum of voltage drops across the right hand side loop must be $0$ by Kirchoff's loop rule. So we have
$$V_{R_2}+V_C=0$$
$$-R_2i-\frac qC=0$$
$$\dot q=-\frac1{R_2C}q$$
So you can see that the time constant only depends on $R_2$. As mentioned in the comments, you can also understand this by noting a $0$ potential drop across the switch. For example, put in a resistor on the branch with the switch and see how coupling between the two sides of the circuit form. Then notice what happens to the coupling when that resistance goes to $0$ like you have in your circuit.
Contrast your case to when the switch is open. Then we only have a loop around the entire circuit, and a similar derivation will show that the time constant depends on both resistances. 
$$V_{R_1}+V_{R_2}+V_C+V_b=0$$
$$-R_1i-R_2i-\frac qC=-V_b$$
$$\dot q=-\frac{q/C-V_b}{R_1+R_2}$$

I'm guessing that it has something to do with how the current changes as time progresses, since the capacitor will slowly allow less and less current to "pass" through as it charges.

Actually, when the switch is closed the capacitor will discharge, not charge. This is because there is no potential difference across the ends of the $R_2\text -C$ branch of the circuit, so there isn't anything to drive the charging of the capacitor. Additionally, the time constant here is, well, constant. It doesn't depend on what the current is in various parts of the circuit or on how much charge is on the capacitor.

As an aside:

The only issue is that when the switch is closed, the circuit becomes parallel

Sure, there are two parts of the circuit that must be experiencing the same potential drop (which is $0$), but I would stay away from thinking about entire circuits as a "parallel circuit". It's best to just think in terms of components (or collections of components) as "in parallel". For example, when the switch is closed $R_2$ and $C$ are in series, so saying the entire circuit is parallel is a little misleading.
