What is the reason for restricting the Special Linear Group $SL_n(\mathbb{C})$ to $\det(A)=+1$? What would be the practical consequence of including $\det(A)=-1$?
Someone else answered before with some hints, that it will not be a subgroup if negative determinant, but he removed his answer. Anyway, in the comments to that answer, I wrote the following, for $A \in SL$, invertibility requirement gives $\pm 1 = \det(I) = \det(AA^{-1}) = \det(A)\det(A^{-1}) = \frac{\det(A)}{\det(A)}$. This leads to a contradiction for negative determinant, since $-\det(A) = \det(A) \implies \det(A) = 0$. This means that $SL_n(\mathbb{C)}$ is only a subgroup when restricting to positive determinants. Is this correct?
Your claim at the bottom is not correct. The orthogonal group $O_n(\Bbb C) := \{M \in GL_n(\Bbb C) \mid \det(M) = \pm 1\}$ does indeed form a useful multiplicative group. And so does the unitary group $U_n(\Bbb C) = \{M \in GL_n(\Bbb C) \mid |\det(M)| = 1\}$.
Your calculation goes wrong when you state $\pm 1 = \det(I)$. But $\det(I) = 1$, never $-1$.
$O_n$ and $U_n$ preserve undirected volume, but not directed volume. In particular, the actions of $O_n$ and $U_n$ do not preserve orientation. $SL_n$ preserves orientation and directed volume, and this is often important. And, as has been noted in the comments $O_n$ is not connected, while $SL_n$ is. But $U_n$ is also connected.