$\DeclareMathOperator{\GL}{GL}\GL_n(\mathbf Z)$ is a multiplicative group, by definition: it is the set of invertible matrices with coefficients in $\mathbf Z$.

The problem is that it's not what you seem to think – the set of matrices with a non-zero determinant. In general, for any (commutative) ring $A$, $\GL_n(A)$ is the set /group of invertible matrices, i. e. the matrices with determinant invertible in $A$.

In the case $A=\mathbf Z$, this means the matrix has determinant $\pm1$.

As your title suggests, ${\rm GL}(n, \mathbb{Z})$ is indeed a group. It consists of those integer matrices with non-zero determinant whose inverses are also integer matrices ( and such matrices all have determinant $\pm 1$, as others have pointed out).

What is not a group is the set of $n \times n$ integer matrices of non-zero determinant. These have inverses with rational entries, but not usually integer entries.

$GL(n,\mathbb{Z})$ is a group for the multiplication law. One can show that : $$GL(n,\mathbb{Z})=\{A\in\mathcal{M}(n,\mathbb{Z})\textrm{ s.t. }|\det(A)|=1\}.$$ As far as $(A,+,\times)$ is a commutative ring with an identity element for $\times$, $GL(n,A)$ is a group.

In a general way, if you consider a matrix $A \in \mathcal{M}_n(\mathbb Z)$, you have the relation $$A.\mathbf{adj}(A)=\det(A)I_n \tag{1}$$ where $\mathbf{adj}(A)$ stands for the adjugate matrix of $A$. The adjugate matrix $\mathbf{A}$ also belongs to $\mathcal{M}_n(\mathbb Z)$.

The relation (1) allows to prove that a matrix $A \in \mathcal{M}_n(\mathbb Z)$ is invertible if and only if its determinant is an invertible element of $\mathbb Z$, i.e. is equal to $\pm 1$.

Which is a proof of the answer provided by Sheol.

Integers with multiplication do not form a group. For example the 1x1 matrix (2) has an inverse (1/2) which is not integer.