That's a good question you asked.

The Standard Model has as a subgroup the Poincaré group, which is the isometry group of flat spacetime, which is both the (global) spacetime of Special Relativity and is the local spacetime of General Relativity for a pretty narrow definition of local, although we regularly construct Local Inertial Frames which are regions of spacetime that are so approximately flat that the difference is negligible.

The Poincaré group has as a subgroup the Lorentz group, and its generators include rotations about the three spacelike axes of Minkowski (i.e., flat) spacetime.

The other symmetries of the Standard Model are invariant under these rotations.

A distant observer moving transversely relative to a particle observes a tiny rotation of the particle. The particle's fundamental properties do not change under that rotation, which can be arbitarily small.

Rather than descending into group theory to reason about a minimum translation or boost, we can look to the energy-mass-momentum equivalence E_r^2 = (m_0 c^2)^2 + (pc)^2 and focus in on photons so we can ignore the m_0 term. Here we have E = E_r = (pc) = \hbar\omega = hc/\lambda = hf. Photons come at arbitrary frequencies, and we have good blackbody radiators all across our sky. If there were a minimal length scale, we would expect that derivatives of position would incorporate that minimal length, and so we would impose observables on things like the doppler shift, for example if we accelerate in a straight line towards a radiator emitting extremely high energy photons.

So it's not so much "why wouldn't it have" but rather, "Q: does observation or experiment support a minimum scale that is large compared to the Planck scale for spacetime intervals in our universe? A: no"

The answers for minimum length et al. scales that are small compared to the Planck scale are subtler (essentially by definition of the Planck scale, at those scales quantum corrections to account for gravitation become significant, spoiling the observability of short spacetime intervals) but so far still 'no'. For every particle in the universe there can be an (ultrarelativistic) observer who sees the particle wavelength shrink below the Planck length; there is nothing special about this observer -- there is likewise a possible observer who sees the particle at some much longer wavelength -- and the point of relativity is that neither observer is more correct than the other. Furthermore, if the former type of observer manages to see a star ultra-blueshifted and extremely Lorentz-FitzGerald-contracted, that observer does not create a black hole; no event horizon forms (event horizons are a global feature of the causal structure of spacetime[0]), and so we can turn an argument about Planck length into an argument about localized Planck energy: there is nothing obviously special in spacetime about the Planck scale. However, our ultrarelativistic observer will certainly see some very strange stuff courtesy of the Unruh effect: different observers observe different particle counts, and one would expect our observer would see an explosion in the number of particles compared to a more typical observer, and those particles -- like all others -- will also interact gravitationally, and we don't know yet how to decomplicate the picture enough to do useful calculations on them.

[0] different observers may disagree on precisely where the event horizon is, what its shape is, and even the count and energy of the particles just outside the event horizon. but they will all agree that there is an event horizon.