How can I check that the voltage output from this voltage divider is 2.25V?

Welcome to resistive potential dividers, if you load them, they change.

You performed a calculation with R1 and R2 forming the potential divider to find the output voltage. However now you are adding in an extra resistor R3. That means that the lower resistor in the potential divider is now actually R2||R3 (R2 in parallel with R3).

In the case of you schematic example, you now have a bottom resistor in the potential divider of R2||R3 = 500Ohms. This is very different from the value you calculated with in the first place. If you repeat the calculation again, you get:

$$V_o = V_i\times\frac{R_2||R_3}{R_2||R_3 + R_1} = 4.5\times \frac{500}{1500} = 1.5V$$

close to what you measured.

As you make the resistor larger and larger, the affect it has becomes less and less - you can see that from the calculation of R2||R3 - the larger you make R3, the closer to R2 the combined value becomes.

It's worth noting at this point that if you omit R3 and simply connect the multimeter across R2, you will actually have the same issue. A multimeter in voltage mode is basically a very large resistor, so if you connect it to your circuit it will still have a loading effect - in essence it becomes R3. However the multimeter resistance is very large (usually >10MOhm), so it will have a very small affect on your circuit.

Simply remove R3. The multi-meter already has a very high input resistance.

You are correct, that you want to "measure the voltage drop of a resistor". However, R2 is that resistor. You don't need to add anything--just measure the voltage drop across R2.