In lectures we were taught, that $\mathbb{Z}_p\cap\mathbb{Q}=\mathbb{Z}$, but also, that there are invertible elements in $\mathbb{Z}_p$, in particular all elements of the form $a_0+a_1\cdot p+\dots$ for $a_0\neq 0$ are invertible.
That implies that $p-1$ is invertible in $\mathbb{Z}_p$, so wouldn't then $\frac{1}{p-1}\in\mathbb{Z}_p$? Where is my mistake here?
You have to be really careful with notation like $\mathbb{Z}_p \cap \mathbb{Q}$. Intersections are really only a meaningful operation inside a fixed ambient set, and depend on a choice of embedding in that set. In this particular case there happens to be a meaningful choice of such an ambient set, namely $\mathbb{Q}_p$. Inside of $\mathbb{Q}_p$ we in fact have
$$\mathbb{Z}_p \cap \mathbb{Q} = \mathbb{Z}_{(p)}$$
which is the localization of $\mathbb{Z}$ away from $p$; explicitly, this is the subring of $\mathbb{Q}$ consisting of fractions $\frac{a}{b}$ where $p \nmid b$. Every $b \in \mathbb{Z}$ such that $p \nmid b$ turns out to be invertible in $\mathbb{Z}_p$. So arguably your lectures were wrong.
In general one cannot make sense of the notation $R \cap S$ for, say, two rings $R$ and $S$ (restricting our attention to rings for simplicity). We need to specify an ambient ring containing both $R$ and $S$, and the answer depends on how we choose to embed $R$ and $S$ into this ambient ring.
$\mathbb{Q}$ is special here because it embeds uniquely into any ring that contains it. The situation is worse for other rings. For example if $p \equiv 1 \bmod 4$ then there is an element $\alpha \in \mathbb{Z}_p$ satisfying $\alpha^2 = -1$. So one might be tempted to ask for the intersection $\mathbb{Z}_p \cap \mathbb{Q}(i)$, since $\mathbb{Q}(i)$ actually embeds into $\mathbb{Q}_p$ in this case. But there are two different such embeddings, and one of them will send $i$ to $\alpha$ while the other will send $i$ to $-\alpha$. So it's misleading to claim that, for example, "$i$ is an element of $\mathbb{Z}_p$" in this case, because it's not possible to unambiguously pick one of the two roots of $-1$ to call $i$.
Here is a fiddly comment. Strictly speaking, in set-theoretic foundations every mathematical object is just a set, so for any two mathematical objects $X$ and $Y$ whatsoever we can ask for their intersection $X \cap Y$. But this is not a meaningful operation to perform on mathematical objects; it depends extremely delicately on how we choose to encode mathematical objects as sets. Arguably it is a huge mark against set-theoretic foundations that they allow us to contemplate this meaningless operation. For example, in set theory, the intersection $\mathbb{R} \cap \mathbb{Q}$ is not simply equal to $\mathbb{Q}$ but is probably empty, depending on exactly how we choose to encode real numbers as sets (via Dedekind cuts or Cauchy sequences or something else), and on how we choose to encode rational numbers as sets. This is of course silly; there is a canonical copy of $\mathbb{Q}$ inside $\mathbb{R}$, which happens to be encoded via different sets than the copy of $\mathbb{Q}$ you build from the integers, but this doesn't matter.
For more related discussion you can see, for example, my old blog post The type system of mathematics, and this math.SE question about a comment of Peter Freyd's that points at similar issues.
