A phototransistor with LED and resistance in emitter
The transistor's collector emitter leakage current is enough to build up a voltage across the LED. You can solve the situation by connecting a large'ish resistor in parallel to the LED. That way the leakage current has a path to ground again. When the photo transistor gets biased by the laser, the current should be large enough for a voltage build up across the resistor to light the LED. 

You probably need a series resistor for the LED too, you didn't mention that in your question.







simulate this circuit &ndash; Schematic created using CircuitLab

The mentioned 2k is just an guesstimate (\$2\text{k}\Omega × 100\mu\text{A} = 200\text{mV}\$) which sounds acceptable and in practice may be better than that). Best is to measure the current through the transistor when not being lit and calculate an acceptable resistor value for that current.

The series resistor should limit the current through the LED to maybe 10 or 20mA (\$\dfrac{V^+-V_{LED}}{10\text{mA}} = \dfrac{V^+-2\text{V}}{10\text{mA}}\$). Also check the optocoupler's maximum current to be sure it can handle 10mA.