In the unbelievable scenario where the required acceleration gradients could be achieved, tau leptons would have an accelerator advantage over muons because of their reduced bremsstrahlung and synchrotron radiation losses. There would otherwise be no significant physics advantage except for probing forces that more strongly affect tau leptons than muons or electrons.
Required Acceleration Gradient
To accelerate unstable particles to an energy $E>>mc^2$ without them decaying, the accelerator must have a mean acceleration gradient
$$G \equiv \frac{dE}{dx} > \frac{mc^2}{c\tau}$$
where $m$ and $\tau$ are the mass and lifetime of the particles. This just follows from the requirement that the mean decay distance $\gamma\beta c \tau$ must be greater than the length of the accelerator $L=E/G$, where $\gamma\beta\approx E/mc^2$.
Accelerating muons ($m_\mu=105.6$ MeV, $\tau_\mu=2.2$μs) only requires $G> 0.16$ MeV/m, which is easy and why muon colliders are a definite possibility. Tau leptons ($m_\tau=1.777$ GeV, $\tau_\tau=2.92\times 10^{-13}$s), however, would require an acceleration gradient of about
$$G_{\mathrm{required}} \sim 20\ \mathrm{TeV/m}$$
This is about 6 orders of magnitude higher than the current high-gradient accelerators, and 2 orders of magnitude higher than the most optimistic gradient for possible future plasma accelerators.
If we go beyond optimism, however, the wavebreaking limit for common solid metals (Cu, Al, Be, Fe, …) with carrier densities $n_e\sim 10^{29}\ \mathrm{m^{-3}}$ is actually slightly above the required gradient for tau lepton acceleration:
$$G_{wb}=eE_0=\sqrt{4 \pi \hbar c (m_e c^2) n_e \alpha} \gtrsim 20\ \mathrm{TeV/m}$$
Note that if such a gradient could be achieved, it would sense to use it to reach ultra high energies where tau lepton decays would not be an issue. A $1$ km tau lepton accelerator would produce a $20000$ TeV beam which would travel an average of about a kilometre before decaying.
The required power density is so high, however, that I'd expect such an accelerator to vaporize the first time it was used, which makes it hard to achieve the necessary luminosity.
Linear Radiation Losses
The radiated electromagnetic energy from an linearly accelerated relativistic particle of mass $m$, electric charge $e$, and energy $E$ is
$$\frac{dE_\gamma}{dx} = \frac{2}{3} \frac{e^2}{4 \pi \epsilon_0} \frac{\hbar c}{(m c^2)^2} G^2= \frac{2}{3} \alpha\hbar c \left(\frac{G}{m c^2}\right)^2$$
where $c$ is the speed of light, $\epsilon_0$ is the vacuum permittivity, $\alpha$ is the fine structure constant, and $\hbar$ is the reduced Planck Constant. (See Eq. 5 of "Does an electron accelerated in a linear accelerator lose any energy?".)
This shows that the radiation loss in an linear accelerator will be unimportant unless the gain in energy is of the order of
$$G_\gamma=\frac{\mathrm{rest\ mass}}{\mathrm{classical\ electromagnetic\ radius}}=\frac{mc^2}{\hbar c \alpha / mc^2}=\frac{(mc^2)^2}{\alpha \hbar c}$$
Even for electrons, this is $2\times 10^{8}$ TeV/m, far above any conceivable acceleration gradient, so in practice tau leptons would have no advantage over electrons in terms of linear acceleration radiation losses.
Accelerator Advantages
In the (wildly overoptimistic) scenario where $\gtrsim 20\ \mathrm{TeV/m}$ acceleration could be achieved, tau leptons would have an advantage over muons because of their reduced bremsstrahlung and synchrotron radiation losses.
Any such accelerator would be accelerating the particles through a dense plasma, where bremsstrahlung and direct $e^+e^-$ pair production losses are inversely proportional to the mass-squared of the accelerated charged particle. (See "Energy loss of muons in the energy range 1-10000 GeV".) Tau leptons bremsstrahlung would be a factor of $(m_\mu/m_\tau)^2\sim 0.004$ less than for muons, and $(m_e/m_\tau)^2\sim 10^{-7}$ less than for electrons.
Any steering or focussing of the beam particles would produce synchrotron radiation losses which scale as the inverse fourth power of the particle mass, so tau leptons would have an even greater advantage compared to muons $\left((m_\mu/m_\tau)^4\sim 10^{-5}\right)$ or electrons $\left((m_e/m_\tau)^4\sim 10^{-14}\right)$.
Physics Advantages
About the only direct physics advantage of a tau lepton collider over a muon collider would be if tau leptons are subject to forces not strongly felt by muons or electrons. For example, if at ultra-high energies there are Higgs-like couplings inversely proportional to fermion mass or purely "tauonic forces" that only couple to tau leptons and tau neutrinos. The latter seems unlikely, but similar exclusive "muonic forces" have been hypothesized.
