Well spotted. In a case like this, it's a good idea to check the article's history (using the "View history" link at the top). In the present case, the error was introduced only two days ago by an anonymous user in this edit (which I just reverted).
The goal is to show that if a square submatrix of $[A,\pm A]$ is nonsingular then it has determinant $\pm 1$. Observe that any nonsingular square submatrix of $[A,\pm A]$ is (up to permutation and sign of the columns) a nonsingular square submatrix of $A$ (justified later). Hence since permutation changes determinants by the sign of the permutation, which is $\pm 1$, and changing the sign of a columns also multiplies the determinant by $-1$ for each such column, we observe that this implies that the determinant of every nonsingular square submatrix of $[A,\pm A]$ is $\pm 1$.
Now to justify the claim. Well, if we choose a square submatrix of $[A,\pm A]$, we are choosing a subset of the rows and columns of $[A,\pm A]$. Let's label the columns of $[A,\pm A]$ as $C_1,\ldots,C_n,C_{n+1},C_{2n}$.
By definition $C_i=\pm C_{i+n}$ for $1\le i\le n$, so if the subset of the columns that we choose contains index $i$ and $i+n$ for some $1\le i\le n$, the resulting matrix must have determinant 0 because $C_i$ and $C_{i+n}$ will be linearly dependent no matter what subset of the rows we choose. Thus there is a map identifying the subset of the columns of $[A,\pm A]$ with a subset of the columns of $A$ given by
$$ i\mapsto \begin{cases} i & \textrm{if $1\le i \le n$} \\ i-n & \textrm{otherwise.}\end{cases}. $$
Best Answer
From http://en.wikipedia.org/wiki/Unimodular_matrix : one definition of a unimodular matrix is an invertible matrix of integers whose inverse is also a matrix of integers. So it is required to be square, in order to be invertible.
So, if $M$ is unimodular, then $M^{-1}$ is a matrix of integers$\ldots$ , and its inverse (which is $M$) is also a matrix of integers, so $M^{-1}$ is also unimodular.
(In fact the unimodular matrices form a subgroup of $GL_n(\mathbb{R})$.)
If you're more interested in non-square matrices, then you probably want to consider total unimodularity, as in the link you provided.