Given an arithmetic sequence $\{a_{n}\}$ with a common difference of $\pi$, we can express the $n$-th term as $a_{n}=a_{1}+(n-1)\pi$. This simplifies to $a_{n}=n\pi +a_{1}-\pi$.

Since the common difference is $\pi$, the sine function, which has a period of $2\pi$, will repeat its values every $2\pi$. However, considering the arithmetic sequence's structure, the effective period for $\sin{a_{n}}$ is $T=\frac{2\pi}{\pi}=2$. This means after every two terms, the sine values will repeat.

To ensure the set $S=\{a,b\}$ contains exactly two distinct elements, we need the conditions:
1. $\sin{a_{1}} \neq \sin{a_{2}}$ to guarantee $a$ and $b$ are different.
2. $\sin{a_{1}} = \sin{a_{3}}$ to ensure the pattern repeats every two terms, aligning with the period $T=2$.

By choosing ${a}_{1}=\frac{\pi}{6}$, we can calculate:
- For $a=\sin{a}_{1}=\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$
- For $b=\sin{a}_{2}=\sin\left(\pi+\frac{\pi}{6}\right)=\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}$

Therefore, when adding these two values, we get $a+b=\frac{1}{2}-\frac{1}{2}=0$.

Hence, the correct answer is $\boxed{B}$.