Instead of thinking of these things as "resistors", try thinking of them as "conductors". After all, that's what they do: conduct.
A resistor with resistance \$R\$ is a conductor with conductance \$S=\dfrac{1}{R}\$.
When you provide multiple conductors connecting one point to another, the conductances simply add. What could be more intuitive? When you provide an additional path for current to flow, more total current flows.
Conductors \$S_1\$ and \$S_2\$ in parallel have a total conductance of:
\$S = S_1 + S_2\$
If you want to express \$S_1\$ and \$S_2\$ as resistances \$R_1\$ and \$R_2\$, you get:
\$S = \dfrac{1}{R_1} + \dfrac{1}{R_2}\$
And, if you want to express the total conductance S as a resistance R:
\$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\$
\$R = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}}\$
Which is the usual expression for the total resistance of two resistors in parallel.
Trivia: the unit of conductance (i.e. inverse ohms) is sometimes called the "mho" ('Ohm' backwards), and is written with an upside-down Omega symbol: ℧. The official SI name for this unit is siemens ("S").
So why is this a valid proof for all resistors in parallel
First, you have an error in your question - the equivalent resistance is
$$R_P = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Now, the voltage across the two parallel resistors is what it is regardless of how the voltage comes to be.
However we choose to label that voltage is immaterial, thus, we can arbitrarily label the voltage across the parallel resistors as, e.g., \$v_P\$.
Now, and again, it does not matter how this voltage comes to be, the voltage variable \$v_P\$ is the voltage measured across the parallel resistors when "red" lead is placed on the "\$+\$" labelled terminal and the "black" lead is on the "\$-\$" labelled terminal.
Thus, by Ohm's law, the current through each resistor is
$$i_{R_1} = \frac{v_P}{R_1} $$
$$i_{R_2} = \frac{v_P}{R_2} $$
So, the total current is, by KCL,
$$i_P = i_{R_1} + i_{R_2}$$
and the equivalent resistance is defined as
$$R_P = \frac{v_P}{i_P}$$
thus,
$$R_P = \frac{v_P}{i_{R_1} + i_{R_2}} = \frac{v_P}{\frac{v_P}{R_1} + \frac{v_P}{R_2}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Again, if we replace the two parallel resistors with a resistor of resistance \$R_P\$, the current through the equivalent resistance will be identical to the sum of the currents through the two parallel resistors.
Best Answer
No, R1 and R2 are not in parallel unless the load was 0 Ω (so vout = 0).
Assuming the load is non-zero, but it is not a simple resistive load that can be separately measured, you can get an equivalent resistance by measuring the current through the load and the voltage across it.
$$R_{load} = \frac{vout}{I_{load}}$$
Then, the resistance from the junction of R1 and R2 to ground would be:
$$R_{parallel} = \frac{R1 + R2 + R_{load}}{R1 \times (R2 + R_{load})}$$
Note that if you plug in 0 for \$R_{load}\$ in the above equation, you get the formula for just R1 and R2 in parallel, as stated in the first paragraph.