How do I replace a CdS sensor in a circuit with an ALS replacement?
In the example circuit, R1 has virtually no effect on the speed of the flashing, since it is so much lower in value than R2. Therefore, if you replace R1 with the ALS, changing the light level will not change the speed of the flashing.

And you can't put the ALS where R2 is, since the circuit relies on current flowing both ways through R2, but the ALS won't allow reverse current to flow.

You're basically going to have to select new component values for the circuit based on how you want it to behave: How fast do you want the LED to blink, what sort of duty cycle do you want?

The existing circuit blinks with a period of about 0.65 seconds. The ALS delivers roughly 0 - 10 mA of current for light levels betweeen 0 and 1000 lux.

As an example of the kind of calculations you need to go through, let's say you replace R1 with the ALS, and you want the minimum on-time of the LED to be 0.1 seconds at 1000 lux. This means that the capacitor much charge from 1/3 Vcc to 2/3 Vcc in 0.1 seconds with a 10 mA current. Since Vcc is 3V, the &Delta;V is 1V.

$$C = \frac{I \cdot t}{\Delta V} = \frac{10 mA \cdot 0.1 s}{1 V} = 1000 \mu F$$

Now, R2 is going to control the off-time of the LED, but we can't allow this resistor to drop more than about 0.5V during the on-time, so it can't have a value more than about 0.5V/0.01A = 50&Omega;. The nearest standard value is 47&Omega;. This means that the off-time will be

$$T = 0.693 \cdot R \cdot C = 0.693 \cdot 47\Omega \cdot 1000 \mu F = 0.033 s$$

... which may be too short to see.

As you can see, the design is becoming unworkable with this set of assumptions.

Also, there are issues with your choice of a 3V coin cell as a power supply. The current draw will be excessive, making the battery life extremely short. Also, is the 555 even specified to work down to 3V in the first place?

Bottom line: You're going to have to select a different device, and perhaps even a different circuit, to achieve what you want to do.