The question is empty since the commonest presentation of the Ehrenfest process is that it is a Markov chain $(S_t)_t$ on the set $L=\{0,1,\ldots,n\}$, whose transition probabilities are, for every $i$ in $L$, $p_{i,i+1}=(n-i)/n$ and $p_{i,i-1}=i/n$ (and not what you wrote).
This Markov chain $(S_t)_t$ is a birth-and-death chain and, as such, $S_{t+1}=S_t+\xi_{t+1}$ for every $t$, where $\xi_{t+1}$ depends on $(S_k)_{k\leqslant t}$ through $S_t$ only, with the distribution one can guess.
The result you mention at the end of your post refers to something different, which is the fact that the process $(S_t)_t$ is a projection of a simpler process with values in a hypercube $K=\{0,1\}^I$ where $I$ is any set of size $n$. To wit, consider the simple random walk $(X_t)_t$ on the graph with vertex set $K$ and the usual edge set. Thus, all the coordinates of $X_{t+1}$ are those of $X_t$ except one, and the exception is chosen, randomly uniformly in $I$. If $S_t$ counts the number of ones amongst the coordinates of $X_t$, one recovers the Ehrenfest process with values in $L$. And in fact, this $K$-valued process is the one the physicists Paul and Tatiana Ehrenfest introduced as a toy model to explain the macroscopic irreversibility of some microscopically reversible dynamics...
The funny thing is that, by the Ehrenfests' construction, $(S_t)_t$ can be realized from $(X_t)_t$ as $S_t=u(X_t)$, for the Markov chain $(X_t)_t$ on $K$ above and the non injective function $u:K\to L$ one can imagine, and that $(S_t)_t$ is still a Markov chain although, in general, such a non injective lumping of a Markov chain does not produce a Markov chain but rather what is called a hidden Markov chain. But a sufficient condition on the lumping to produce a Markov chain is well known and the Markov process $(X_t)_t$ on $K$ and the Ehrenfest function $u$ from $K$ to $L$ happen to satisfy this condition!