If the light is bouncing (off the mirrors) in the same direction as B's spaceship is moving, A would see exactly what B does: a single beam of light bouncing off each mirror alternately, retracing its own path over and over again.
Remember that the mirrors are moving. So when the light beam travels from the rear mirror to the forward mirror, observer A would actually see a light beam having to catch up to a receding mirror. Similarly, when the light beam travels from the forward mirror to the rear mirror, observer A would see the mirror catching up to the light. This means that according to observer A, the light beam travels further each time it goes forward than when it goes backward, as this image shows:
Even the two halves of the light beam's trip are not the same length according to A, so clearly A and B have to measure different intervals for at least one of those halves (and in fact both).
Quantitatively, suppose the relative speed of A and B is $v$ and the distance between the mirrors (as seen by A) is $\Delta x_A$. On the forward trip of the light beam, as observed by A, the position of the light beam is described by $x_\text{light} = ct$ and the position of the forward mirror is described by $x_\text{mirror} = \Delta x_A + vt$. The time it takes for the light to reach the mirror is obtained by setting these equal to each other:
$$\Delta t_\text{forward} = \frac{\Delta x_A}{c - v}$$
On the backward trip of the light beam, observer A sees $x_\text{light} = -ct$ and $x_\text{mirror} = -\Delta x_A + vt$, so
$$\Delta t_\text{backward} = \frac{\Delta x_A}{c + v}$$
Adding it up, you get a total round-trip time of
$$\Delta t_{A,\text{total}} = \frac{2c\Delta x_A}{c^2 - v^2}$$
Now, suppose you want to find the relationship between $\Delta x_A$ and $\Delta x_B$, the proper distance (i.e. as seen by B) between the mirrors. Hopefully it should be clear that if you look at this from B's perspective, you get
$$\Delta t_{B,\text{total}} = \frac{2\Delta x_B}{c}$$
If you believe the time dilation formula (and it sounds like you do), you can write
$$\Delta t_A = \frac{\Delta t_B}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and now combining the last three equations leads you to
$$\Delta x_A = \Delta x_B\sqrt{1-\frac{v^2}{c^2}}$$
Basically, time dilation is able to account for part of the factor of $\bigl(1 - \frac{v^2}{c^2}\bigr)$ difference between $\Delta t_{A,\text{total}}$ and $\Delta t_{B,\text{total}}$, but not all of it. We have to attribute the rest to length contraction.
Since the only constant in all of this is the speed of light, the only way acceptable to all observers is to measure a distance using c as a yardstick. So... imagine that A sees a beam of light start at the 'rear' of B's spaceship and make its way forward to the 'front' of the spaceship.
Actually, that's not the best way to go about measuring distances, for exactly the reason I described above. As you saw, if you time a light beam traveling from the back of the spaceship to the front (or from a rear mirror to a forward mirror), the time you will actually measure is $\Delta t = \frac{\Delta x}{c - v}$, not $\Delta t = \frac{\Delta x}{c}$ as you thought. (Of course, in a reference frame where the distance being measured is at rest, this works fine since $v = 0$.)
The easiest and recommended way to measure the distance of a moving object is by sitting still at a point and recording the times when the front and back of the object pass you. Once you have the time difference, you can determine the length of the object in your reference frame by $\Delta x = v\Delta t$, where $v$ is the object's speed relative to you. Since boosts between different reference frames "mix" time and space, it's easiest to keep your spatial coordinate fixed when you're measuring time, and vice-versa.
You find this hard to refute because your friend is correct in one sense: the MM experiment did not prove Einstein's second postulate of relativity, namely that the speed of light is constant for all inertial observers.
Recall that the Michelson-Morley experiment was designed to detect motion relative to an aether, or material medium for light. If your experiment on an open train carriage measured the speed of sound, then you would indeed measure different speeds along and across the carriage. So the MM experiment cast serious doubt on the notion of an aether.
Now, it was well known that Maxwell's equations did not keep their form under Galilean transformations between inertial frames. This was thought to be fine because the notion of a medium for light was believed before the MM experiment, so that the wave equation for light should transform in the same way as the wave equation for sound between inertial frames.
So along comes Einstein and says, given there's no medium, let's see what happens to our physics if we assume that Maxwell's equations keep their form under a transformation between inertial frames. He postulated therefore that the speed of light would be measured to be the same for all inertial observers and concluded that (1) the transformation group was the Lorentz, not the Galilean group and (2) the time measured between two events would in general depend on the observer. (1) was already know at the time of Einstein's 1905 paper, (2) was radical.
So the MM experiment motivated the assumed Lorentz covariance of Maxwell's equations and thus the new relativity postulate that the speed of light would be measured to be the same by all inertial observers.
The second relativity postulate therefore comes into play only when we compare the light speed measured by different inertial observers. Someone observing a light source on your train would notice a very different transformation law from the approximate Galilean transformation law that would describe the ping-pong ball velocity transformation to an excellent approximation.
Best Answer
This has absolutely nothing to do with relativity. Of course A says that his own clock ticks at one second per second, as does B. This is as true in Newton's world (or in Aristotle's) as in Einstein's. The key question for relativity is: How fast does A say that B's clock ticks? And your analysis does not address this at all.