You could add them in series - if there was nothing in parallel with R2.
Like you said, if you could add them in series, Vin = Vout which is clearly wrong because it disregards the voltage drop across R1. Essentially you've replaced R1 with a short.
In this circuit, R1 is in series with parallel combination of R2 and C1. R1 is not in series with R2 alone, so you can't combine them.
EDIT: @WhatRoughBeast makes an excellent point in that you can combine the resistors by transforming the circuit into a Thevenin Equivalent. You could also use a Norton Equivalent and replace V, R1, and R2 with a current source and a resistor in parallel, however in my short experience studying EE I find that Thevenin is used far more than Norton.
Thevenin's Theorem allows us to simplify any linear network consisting of voltage sources, current sources, and only resistances into a single Thevenin voltage source \$V_{TH}\$ (not to be confused with the threshold voltage of a MOS transistor) and a Thevenin resistance \$R_{TH}\$.
For your circuit (you can follow along with the schematic that @WhatRoughBeast posted) you ignore C1 and calculate the Thevenin Equivalent of V1, R1, and R2. There are two steps:
- Find \$V_{TH}\$:
The open circuit voltage is simply the output voltage at R2, which is the voltage divider equation. Thus \$V_{TH} = \frac{R_2}{R_1 + R_2} V_1 \$.
- Find \$R_{TH}\$:
Short the voltage source V1 and look at the equivalent resistance through the output terminals. You just have \$R_1 || R_2\$, so \$R_{TH} = \frac{R_1 R_2}{R_1 + R_2}\$.
Training yourself to see opportunities to use Thevenin Equivalents will come in very useful as an EE.