Troubleshooting my own PCB
\$V_f\$ for the laser diode is 2.5V.  Battery voltage of 3.0 leaves 0.5 dropped across R2.  That's \$ \frac {0.5}{400}= 1.25 mA\$. The laser is rated for about 33mA.
Your circuit is limiting the current to the laser diode too much.
Then there's the current rating of the battery to consider.  The CR2025 isn't rated to deliver the current you need, and probably struggles to deliver the current your circuit consumes. (Just a couple of milliamperes is a heavy load for the CR2025.)
Now, take a look at the 555.
Its datasheet says the NE555 can operate on 4.5V to 16V, and that it will draw several milliamperes even at low voltage:



You need a higher operating voltage for the 555.
You need a battery capable of delivering the required current.
You need to re-calculate R2, taking into account the voltage at the output pin of the 555 (pin 3) and the current the laser diode needs (at least 24 mA, but no more than 40 mA.)