Does the mass of a bicycle directly affect stopping distance?

The answer is a little more nuanced than a simple yes or no, but for most cyclists stopping distance will increase with mass. Allow me to explain how:

We can use the work-energy theorem to write down the distance $x$ an object traveling at velocity $v$ will require if a force $F$ is applied opposite to $v$: $$\begin{align} W &= \Delta K \\ -Fx &= 0 - {1 \over 2}mv^2 \\ x &= \frac{m v^2}{2F} \tag{general stopping distance}\\ \end{align}$$

So you can see in general stopping distance is proportional to mass. However, for objects that use friction (like cars and cycles) between the object and ground to stop, the maximum force you can get from friction is also proportional to the object's mass: $F_{max} = \mu m g$ where $\mu$ is the coefficient of friction and $g$ is the gravitational acceleration. Putting the maximum force into the stopping distance yields the minimum stopping distance:

$$x_{min} = \frac{v^2}{2\mu g} \tag{minimum distance}\label{x_min} $$

This minimum stopping distance is mass independent. When you apply your brakes, (usually) a caliper applies a force to the wheel. This force depends on how hard you brake, and the location of the caliper, and lots of other engineering specifics. What it doesn't depend on is the total mass of the object, so $m$ will not cancel out of the stopping distance.

summary

1. All other things being equal (including how hard you apply brakes), stopping distance is proportional to mass.
2. There is a minimum attainable stopping distance, which is independent of mass.

Edit:

An example might clear up some confusion in the comments. Imagine two cyclists, "tiny Tim" and "big Bob". Both are riding identical bikes but Bob has more mass than Tim. They approach a stop sign and wish to come to a complete stop with the same initial velocity: Since Bob has more mass he will have to apply his brakes harder than Tim does, i.e. generate a larger force, if he wishes to stop in the same distance. However, Bob's extra weight generates more available friction with the ground, and so his maximum available stopping force is larger than Tim's.
If both need to stop in the minimum amount of distance, they should apply a braking force up to the maximum allowable by friction between the ground and wheels. Any more and they risk wheel slippage which will raise their braking distance. Thus Bob's minimum stopping distance is the same as Tim's because his max available force is proportionally larger (e.g. if he's twice as massive, he has twice the maximum braking force before his wheels slip).

That depends on what the limiting factor is in the stopping distance before changing the mass.

If you could say that the limiting factor was the coefficent of friction between the tires and the road (braking hard to the point of slipping), then that would remain relatively unchanged with increased mass, and you'd have similar braking performance.

But I suspect that more bicycle stops are limited by either brake effectiveness or technique. Increased mass will make brake effectiveness worse (similar stopping force, but increased kinetic energy to dissipate). I don't know enough to speculate if poor technique is helped or exacerbated by a heavier bike.

The force that the brakes can apply to the wheel depends only on the friction characteristics of the brake-wheel interface, the strength of the rider's fingers and the mechanical advantage afforded by the braking system. All of that is independent of the mass that's being carried by the bike.

As soon as that, rather than the friction between the tyres and the road, becomes the limiting factor, increased weight will increase stopping distance. To take an absurd example, if you're cycling around with a thousand tons of neutron star in your backpack, nothing you do with your brakes is going to slow you down by any appreciable amount.