Why do gauge bosons/leptoquarks not mediate proton decay in the Pati-Salam model?
The answer to your question requires some knowledge on group theory and tensor analysis, but I will try to make it as simple as possible with out going into too much of technicalities.
Your question consists of basically two completely disjoint parts, they are:
why the gauge bosons(leptoquarks) of Pati-Salam Group do not mediate proton decay. Which is a correct statement.
why the scalar bosons(leptoquarks) of Pati-Salam Group do not mediate proton decay. This statement is not always true btw.
First thing to note is, leptoquarks are the particles that carry lepton (L) and baryon (B) numbers and the second thing to note is, for proton decay processes, $B$ and $L$ numbers have to be violated. Now let me answer them separately.
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01=> The gauge bosons corresponding to the $SU(4)$ group is 15-plet. Under the sub-group $SU_{c}(3)\times U_{B-L}(1)$, the decomposition is $15=1+3+\overline{3}+8$. Of these four, two of them has non-zero $B-L$ charge, $3$ and $\overline{3}$ which are $4/3$ and $-4/3$ respectively. So $3$ and $\overline{3}$ are leptoquarks.
Now, if one writes down the Lagrangian containing these leptoquark gauge bosons (to avoid technicalities I am not writing down the Lagrangian here), it is easy to see that, this part of the Lagrangian not only conserves $B-L$ number but also $B+L$, which means, $\Delta B=0$ and also $\Delta L=0$, i.e, separately $B$ and $L$ numbers are conserved and so proton decay is not possible by the leptoquark gauge bosons.
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02=> Now, lets talk about proton decay by the scalar leptoquarks. This is completely model dependent. You can build a model where proton decay cannot happen by the scalars and you can also build models where proton decay is permitted.
Let me tell you the reason. If you are already familiar with Standard Model or QCD, then you know, writing down the Lagrangian involves the $SU_{c}(3)$ invariant tensor $\epsilon_{ijk}$ (called levi-civita tensor). $i,j,k=1-3$ (or for simplicity, red,green,blue) as the group is $SU_{c}(3)$ and the indices are color indices. Such terms in the Lagrangian introduces Feynman diagrams that involves transition from one colored quark to another colored quark.
Now, in Pati-Salam group, you have $SU_{c}(4)$, so writing down the full Lagrangian for the scalars may involve the invariant tensor $\epsilon_{ijkl}$ with $i,j,k,l=1-4$ (to avoid technicalities I am not writing down such terms here). As I mentioned before, whether the Lagrangina will contain this factor or not, that depends on what kind of scalars you are introducing in your model. The important thing to note is, $1-3$ for usual quark color indices but $4$ is the fourth color for leptons. So, such terms has interactions involving quarks and leptons. The point is, if such term exists, it is clear that Feynman diagrams will be allowed, which will involve transitions between quarks and leptons that will allow $B$ and $L$ non-conservations and will lead to proton decay.
While the answer provided above is correct, there's maybe simple / less technical way to see this. I put it here hoping it would help other curious minds.
In any gauge theory based on a simple non-abelian group, gauge bosons mediate interactions between (i.e., connect) particles belonging to the same multiplet (example: in the SM, the fact that charged lepons and neutrinos belong to the same $SU(2)_L$ doublet means that there's a gauge boson which connects them: the $W^\pm$). So we would reasonably expect bayron number violating transitions (quark $\to$ lepton) leading to proton decay in any model the combines leptons and baryons in the same multiplet.
In Pati-Salam (PS), the $SU(4)$ fundamental representation indeed (partially) unifies quarks and leptons: in the $4-$dimensional multiplet we have 3 quarks, corresponding to the three colors, and 1 lepton. (I ignore the $SU(2)_{L,R}$ for now.) So we could expect proton decay since there should be vector lepto-quarks connecting the lepton to the quarks. However, for this very specific case, this kind of interactions will not lead to proton decay because baryon number is a gauged symmetry; $SU(4)$ contains $U(1)_{B-L}$ and preserves it. Remember that we have the chain: \begin{eqnarray} SU(2)_L \times SU(2)_R \times SU(4)&\to& SU(2)_L \times SU(2)_R\times U(1)_{B-L} \times SU(3)_C\\ &\to& SU(2)_L \times U(1)_Y \times SU(3)_C \\ &\to& U(1)_{em} \times SU(3)_C \end{eqnarray}
So baryon and lepton number violating transitions are absent (or, at least, very suppressed depending on the PS variant).
For proton decay mediated by scalar leptoquarks the situation is different (as SAS points out in the previous answer) and we cannot make model-independent claims. But let's say that in principle you can usually avoid this kind of proton decay processes or make them very small (unless you want it!).
Bonus 1:
In more elaborate GUTs where the unification is complete (based on $SU(5)$, $SO(10)$, $E_6$, etc.), quarks and leptons appear in the same multiplet(s) (e.g., in the case of $SO(10)$, all leptons and quarks of the SM are in a ${16}-$dimensional multiplet); proton decay is then unavoidable. We can estimate the lifetime of the proton as:
$$\tau_\mathrm p = \frac{M_\text{GUT}^4}{\alpha_U^2 M_\mathrm p^5}\,,$$
where $M_\mathrm{GUT}$ is the mass scale of the gauge bosons inducing proton-decay, $M_\mathrm p\approx 1\,\mathrm{GeV}$ is the proton mass, and finally $\alpha_U$ is the gauge coupling strength at unification scale (typically, $\alpha_U\approx 1/20$ ). To comply with current bounds on the lifetime of the proton, $\tau_{\mathrm{exp}}\gtrsim 10^{34}$ years, the scale of these new gauge bosons (and therefore of the breaking of the GUT group) must be $\gtrsim 10^{14 - 16}$ GeV.
Bonus 2:
We know that PS $\subset SO(10)$, therefore $U(1)_{B-L} \subset SO(10)$, so why doesn't the argument above apply to $SO(10)$? why do we have proton decay there?
The fundamental representation which contains all the SM fermions + a new iso-singlet neutrino is $\mathbf{16}$, which at the level of PS this becomes: $$ \mathbf{16}\to (\mathbf{4},\mathbf{1},\mathbf{2}) \oplus (\mathbf{\overline{4}},\mathbf{2},\mathbf{1}) $$ which are but the basic representations of the PS model. However, $SO(10)$ has 45 generators (gauge bosons), whereas PS has 'only' 3+3+15=21. So there are 24 extra gauge bosons which belong only to SO(10), we can see this from the decomposition of the adjoint of $SO(10)$: $$ \mathbf{45}\to (\mathbf{1},\mathbf{3},\mathbf{1}) \oplus (\mathbf{1},\mathbf{1},\mathbf{3}) \oplus (\mathbf{15},\mathbf{1},\mathbf{1}) \oplus \color{red}{(\mathbf{6},\mathbf{2},\mathbf{2})} $$
These correspond to $SU(2)_L$, $SU(2)_R$, $SU(4)$ and the extra stuff, resp.
In PS, with two fermions and a vector, we can only make (focusing on the $SU(4)$ part) $\mathbf{4}_F.\mathbf{{4}}_F^\star.\mathbf{15}_V$ products which, as argued above, do not lead to proton decay. However with the extra bosons, we can make: $\mathbf{4}_F.\mathbf{\overline{4}}_F^\star.\mathbf{6}_V$ ... leading to proton decay.