Why can we detect gravitational waves?

The short answer is that waves that are "in the apparatus" are indeed stretched. However the "fresh waves" being produced by the laser are not. So long as the "new" waves spend much less time in the interferometer than it takes to expand them (which takes roughly 1/gravitational wave frequency), then the effect you are talking about can be neglected.

Details:

There is an apparent paradox: you can think about the detection in two ways. On the one hand you can imagine that the lengths of the detector arms change and that the round-trip travel time of a light beam is subsequently changed and so the difference in the time-of-arrival of wavecrests translates into a phase difference that is detected in the interferometer. On the other hand you have the analogy to the expansion of the universe - if the arm length is changed, then isn't the wavelength of the light changed by exactly the same factor and so there can be no change in the phase difference? I guess this latter is your question.

Well clearly, the detector works so there must be a problem with the second interpretation. There is an excellent discussion of this by Saulson 1997, from which I give a summary.

Interpretation 1:

If the two arms are in the $x$ and $y$ directions and the incoming wave the $z$ direction, then the metric due to the wave can be written
$$ds^2 = -c^2 dt^2 + (1+ h(t))dx^2 + (1-h(t))dy^2,$$
where $h(t)$ is the strain of the gravitational wave.

For light travelling on geodesic paths the metric interval $ds^2=0$, this means that (considering only the arm aligned along the x-axis for a moment)
$$c dt = sqrt{(1 + h(t))}dx simeq (1 + frac{1}{2}h(t))dx$$
The time taken to travel the path is therefore increased to
$$tau_+ = int dt = frac{1}{c}int (1 + frac{1}{2}h(t))dx$$

If the original arm is of length $L$ and the perturbed arm length is $L(1+h/2)$, then the time difference for a photon to make the round trip along each arm is
$$ Delta tau = tau_+ - tau_- simeq frac{2L}{c}h(t)$$
leading to a phase difference in the signals of
$$Delta phi = frac{4pi L}{lambda} h(t)$$

Interpretation 2:

In analogy with the expansion of the universe, the gravitational wave does change the wavelength of light in each arm of the experiment. However, only the waves that are in the apparatus as the gravitational wave passes through can be affected.

Suppose that $h(t)$ is a step function so that the arm changes length from $L$ to $L+h(0)/2$ instantaneously. The waves that are just arriving back at the detector will be unaffected by this change, but subsequent wavecrests will have had successively further to travel and so there is a phase lag that builds up gradually to the value defined above in interpretation 1. The time taken for the phase lag to build up will be $2L/c$.

But then what about the waves that enter the apparatus later? For those, the laser frequency is unchanged and as the speed of light is constant, then the wavelength is unchanged. These waves travel in a lengthened arm and therefore experience a phase lag exactly equivalent to interpretation 1.

In practice, the "buildup time" for the phase lag is short compared with the reciprocal of the frequency of the gravitational waves. For example the LIGO path length is about 300 km, so the "build up time" would be 0.002 s compared with the reciprocal of the $sim 100$ Hz signal of 0.01 s and so is relatively unimportant when interpreting the signal.