Is any relation which contains only one ordered pair transitive?

The transitive condition is true vacuously. That's like saying "All the women in the car are on fire" is true, when a man is in the car alone.

Yes, it is a transitive relation, vacuously so. That is, there are no counter examples in the relation that violate transitivity.

Transitivity requires that

If $(P)$: $(i)$ $(a, b) \in R\;$ AND $(ii)$ $(b, c) \in R$, (conditions)

THEN $(Q)$: it must follow that $(a, c) \in R$ (consequent)

Since $(P)$, the conditions (i) and (ii), will never both be realized/satisfies since the only element in $R$ is $(1, 2)$, we have that the implication $(P) \implies (Q)$ is vacuously true.

NOTE: We can equivalently define transitivity as a property that HOLDS UNLESS there exists a case (counterexample) for which both conditions in $(P)$ are met, but the consequent $(Q)$ is false (does not hold.)

$R$ is transitive in an empty sense, because in a transitive relation, we want the following $ \forall x,y,z\in A, (x,y)\in R \wedge (y,z)\in R \Rightarrow (x,z)\in R $, but this doesn't guarantee the existence of $3$ pairs in $R$. For simplicity, let's write $ \alpha =[(x,y)\in R \wedge (y,z)\in R], \beta =[(x,z)\in R] $. So what we want is that $ \alpha \Rightarrow \beta$, but if $ \alpha $ has a truth value of $0$, then according to the truth table of implication, then $\beta$ will always have a truth value of $1$, which means that a relation $R$ is transitive if $ \alpha$ implies $ \beta $ in $R$.

For example in this case, only $(1,2)$ is in $R$ (meaning $ \alpha $ has a truth value of $0$, because we cannot find $2$ distinct pairs in $R$), and therefore, $ \beta $ has a truth value of $1$, therefore, we found that $ \alpha \Rightarrow \beta$, and therefore, $R$ is transitive.