Though beta positive decay's have been observed, in which a proton decays into a neutron, positron and an electron- neutrino; the why is it not the same as the exotic proton decay which is hypothetical and hasn't been observed?
Protons and neutrons are very similar particles. Although they have different charges, as far as the strong force is concerned they are almost identical. So changing a proton into a neutron and back isn't considered decay because you aren't changing the number of nucleons. More precisely the baryon number remains constant.
If we describe the proton and neutron as a bound state of three quarks$^1$ then a proton is two up and one down quarks while a neutron is two down and one up quarks. Interchanging them requires changing an up quark to a down quark and vice versa, which happens by emission/absorption of an electron or positron. The number of quarks doesn't change and the process is reversible.
The process we normally describe as proton decay is altogether more radical. There are actually several different possible mechanisms for proton decay, but they all involve the proton disappearing completely leaving behind just a positron and two photons. This has to involve the creation of a hypothetical and so far unobserved particle called an X boson. The Standard Model does not include the X boson so as far as the Standard model is concerned the proton cannot decay.
$^1$ Caution! Not literally true!
Protons and neutrons belong to an isospin doublet, with baryon number one. Baryon number is a conserved quantity, a law coming from experimental observations. Here is the representation where the proton and the neutron belong, the baryon octet:
These particles decay to the lowest mass representative of baryons , the proton, and the proton is considered stable.
Beta + decay is allowed if there is enough energy, but again , baryon number is conserved.
With the discovery of the quarks, the standard model is consistent by giving 1/3 baryon number to each quark. The question of proton itself decaying, is answered in the negative. It is all due to the baryon number conservation law which is fulfilled in the mathematics of the standard model..
There exist though models beyond the standard model, and there one can write diagrams and compute probabilities for a proton to decay, for example in the diagram below neither baryon number nor lepton number rules of the standard model are obeyed.
The x is a leptoquark in this model, and they are searching for it at the LHC.
There are running experiments that check these proposed extensions of the standard model and they have set limits on proton decays.
A 2014 result with 260kT·yr of data, searching for decay to K-mesons set a lower limit of 5.9 × 10^33 yr, close to a supersymmetry (SUSY) prediction of near 10^34 yr .
Protons have smaller masses than neutrons. It is energetically favorable for a free neutron to decay into a (proton, electron, anti-neutrino) but not possible for a free proton to decay into a (neutron, positron, neutrino). Inside a nucleus, however, the latter process is possible (positive beta decay). In either case the number of baryons remains the same. The exotic proton decay leaves no baryons what-so-ever.