Notation: Let $R$ be a PID with field of fractions $K$ and $A$ a quaternion algebra over $K$. Take an order $\mathcal O \subset A$, i.e. an $R$-subalgebra that is a finitely generated $R$-module. Denote by $\operatorname{Tr}$ the (reduced) trace on $A$. We have the dual $\mathcal O^* = \{ x \in A : \operatorname{Tr}(x\mathcal O) \subset R \}$ and the different $D(\mathcal O) = (\mathcal O^*)^{-1} = \{x \in A : x \mathcal O^* \subset \mathcal O \}$.
In Vigneras' book, Lemma 4.7 says that when $(e_i)$ is a basis of $\mathcal O$, then $\det(\operatorname{Tr}(e_ie_j))R$ is the square of the ideal generated by the reduced norms $N(x)$ for $x \in D(\mathcal O)$. It is proved only in the case where there exists $x \in A$ with $x\mathcal O^* = \mathcal O$. In particular, $\mathcal O^*$ is then invertible, i.e. $D(\mathcal O) \mathcal O^* = \mathcal O^* D(\mathcal O) = \mathcal O$. I am unable to adapt the proof to the general case.
The proof goes like: let $(e_i^*)$ be the dual basis, then this is a basis of $\mathcal O^*$ and when $e_i = \sum a_{ij} e_i^*$ then $\det(\operatorname{Tr}(e_ie_j)) = \det(a_{ij})$. Now take $x \in D(\mathcal O)$ and write $x e_i^* = \sum_j \lambda_{ij} e_j = \sum_{j,k} \lambda_{ij} a_{jk} e_k^*$ with $\lambda_{ij} \in R$, then $N(x)^2 = \det(\lambda_{ij}) \det(a_{jk}) \in \det(\operatorname{Tr}(e_ie_j))R$. For the reverse inclusion, there exists $x \in D(\mathcal O)$ with $x \mathcal O^* = \mathcal O$, and then $\det(\lambda_{ij}) = 1$.
Question: Does the identity $\det(\operatorname{Tr}(e_ie_j))R = N(D(\mathcal O))^2$ really hold when $\mathcal O^*$ is not invertible? Or is it true that $\mathcal O^*$ is invertible for all $\mathcal O$?
My impression is no: with the notations from the proof, there should exist for every irreducible $\pi \in R$ an element $x \in D(\mathcal O)$ with $\det(\lambda_{ij}) \notin \pi R$. Then $x : \mathcal O^*/\pi \mathcal O^* \to \mathcal O/\pi \mathcal O$ is an isomorphism; in particular $D(\mathcal O) \mathcal O^* + \pi \mathcal O = \mathcal O$. But this means that the invariant factors of $\mathcal O / D(\mathcal O) \mathcal O^*$ are all $1$. Similarly for $\mathcal O^* D(\mathcal O)$. I.e. $\mathcal O^*$ is invertible.
Note: if $\mathcal O^*$ is invertible for all $\mathcal O$, then every ideal is invertible. Indeed, for every ideal $I$, $II^* = \mathcal O_\ell(I)^*$ and $I^*I = \mathcal O_r(I)^*$ are then invertible
This does look like an error in Vigneras's book. In fact, for an order $\mathcal O$ in a quaternion algebra $A$, TFAE:
Proof. $1 \implies 2$ is in Vigneras, together with Kaplansky's theorem that an ideal is invertible iff it is locally principal. (Theorem 16.6.1 in Voight's book on quaternion algebras.) $2 \implies 1$ is sketched in my question. Or more precisely: If $\sqrt{\det(\operatorname{Tr}(e_ie_j))R} \subset Nrd(D(\mathcal O))$, then $\det(\operatorname{Tr}(e_ie_j)) \in Nrd(D(\mathcal O))^2 \subset N(D(\mathcal O))$, and as in the question it follows that the invariant factors of $\mathcal O / D(\mathcal O) \mathcal O^*$ are all $1$.
Example: take $A$ to be the Hamilton quaternions over $\mathbb Q$ and $\mathcal O$ generated by $1, 2i, 2j, 2k$. Then $\mathcal O^*$ is the lattice spanned by $\frac12, \frac i4, \frac j4, \frac k 4$ and $D(\mathcal O)$ is spanned by $8, 8i, 8j, 8k$. Clearly, $1 \notin \mathcal O^* D(\mathcal O)$.