To answer your question, you first have to choose a map $\mathbb Q[\mu_{p-1}] \subset \mathbb Q_p.$ To this end, note that $p \equiv 1 \bmod p-1,$ hence $p$ splits completely in $\mathbb Q[\mu_{p-1}]$, and so choosing such an embedding amounts to choosing one prime $\mathfrak p$ above $p$. The embedding is then characterized by the fact that the maximal ideal of $\mathbb Z_p$ pulls back to $\mathfrak p$, while all the other prime ideals dividing $p$ generate the unit ideal of $\mathbb Z_p$.
In order to keep things concrete, it will be easier to think about actual elements (or, if you like, pricipal ideals) rather than prime ideals that are not necessarily principal. To this end, choose a power $h$ such that $\mathfrak p^h$ is principal, say equal to $\pi$. We may then write p^h = \pi \pi', where \pi' is prime to $\mathfrak p$. (The ideal generated by \pi' is the product of the $h$th powers of all the primes other than $\mathfrak p$ which divide $p$.) Then we see that \mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] = \mathbb Z[\mu_{p-1},\frac{1}{(p-1)^h\pi\pi'}].
The elements $p-1$ and \pi' are invertible in $\mathbb Z_p$, while $\pi$ is not (it generates the $h$th power of the maximal ideal of $\mathbb Z_p$). Thus we see that \mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbb Z_p = \mathbb Z[\mu_{p-1},\frac{1}{(p-1)\pi'}], and so its Spec coincides with that of $\mathbb Z[\mu_{p-1}]$ with the prime above $p-1$ and all the primes dividing \pi' (i.e. all but one of the primes over $p$) removed (confirming your claim).