The best place to seek evidence that decay doesn't equal compositness is in particle *creation*. Because if decay meant compositeness, then creation would require you to get the constituents together.

When you bash two nucleons together at high enough energy you get a lot of junk coming out. Some of that junk is lepton particle-antiparticle pairs, and many of them arise from interactions like
$$ q + \bar{q} \to l^- + l^+ \,.$$
This process is called "Drell-Yan". The leptons can be electrons, muons or tauons. Getting muons is experimentally very useful, so this process is sometimes used as a probe of the structure of the nucleon sea. (When you put protons on protons, the anti-quarks have to come from the sea as the valence content is all quark.)

If you have an electron-positron machine at high energy (i.e. the decommissioned SLC or LEP) you can also do
$$ e^- + e^+ \to l^- + l^+ \,,$$
with similar mathematics.

Now, at energies very much over $2m_\mu c^2$, the rate for producing electron-pairs and that for producing muon-pairs is the same, which wouldn't be the case if one were elementary and the other composite (if the muons were composite the chance of having the right bits present would contribute to the production rate so the rate would be lower). Further the rate is in agreement with the *ab initio* predictions from QED for fundamental leptons. Let's take a moment to recall that QED offers the best single agreement between theory and experiment in physics (the electron's g-2).

In addition there are many other interesting predictions from QED about muons (for instance the muon g-2 which is nearly as good a theory-experiment match as that for electrons).