Why does the congruence hold? Let $p$ be a prime number and $\mathbb{Q}(\zeta)$ be the pth cyclotomic number field where $\zeta$ is any primitive pth root of unity.
Writing $$t=b_0+b_1\zeta+...+b_{p-2}\zeta^{p-2} $$ with $b_j \in \mathbb{Z}$ , we get $$t^p \equiv b_0^p+b_1^p+...+b_{p-2}^p \pmod{p\mathbb{Z}[\zeta]}$$
I understand the congruence if I consider this modulo p .

The Binomial Theorem based proof of the Freshmans's Dream $\, (a+b)^p = a^p + b^p\,$ in $\,\Bbb F_p\,$ uses only the ring axioms plus $\,p = 0\,$ and $\,a,b\,$ commute. So the proof will work in any such ring.
Said universally, $\,(x+y)^p = x^p + y^p$ in $\Bbb F_p[x,y]\,$ so it holds in any ring containing $\,\Bbb F_p\,$ by the universal mapping property of polynomial rings, i.e. simply evaluate the polynomoial dream at $\,x = a, y = b\,$ to map it into any ring where $\,p=0\,$ and $\,a,b\,$ commute.
In particular the dream is true in your $\,S = R/pR\,$ by $\,p \in pR\,\Rightarrow\, p=0\,$ in $S$
