If $A$ is an $m\times n$ matrix, then the equation $Ax = b$ has a unique least-squares solution for each $b\in \mathbb{R}^n$ if and only if $\det(AA^T)\neq 0$.
I believe it is true because if A is m x n matrix, then the equation $Ax=b$ has a unique solution for each $b\in\mathbb{R}^n$ only if $A$ is invertible? My logic could be wrong so I am hoping for a second opinion.
Even if we assume that $b\in\mathbb{R}^{m}$, then the statement is false. You can look at the pseudo-inverse which is the solution to the least-squares problem. The pseudo-inverse requires that \begin{align} A^{\mathsf{T}}A \end{align} be invertible; it says nothing about $AA^{\mathsf{T}}$.
An immediate counter-example is any column matrix $A\in\mathbb{R}^{m}$ such that $\Vert{A}\Vert_{2}\neq 0$. In such case, $\operatorname{det}(AA^{\mathsf{T}})=0$ but $\operatorname{det}(A^{\mathsf{T}}A)=\Vert A\Vert_{2}^{2}\neq 0$, and the least-squares problem has a unique solution $$ x^{\star} = (A^{\mathsf{T}}A)^{-1}Ab. $$
On the other hand, you can have a row matrix $A\in\mathbb{R}^{1\times n}$ such that $\Vert A^{\mathsf{T}}\Vert^{2}_{2}\neq 0$ which satisfies $\operatorname{det}(AA^{\mathsf{T}})\neq 0$ but doesn't satisfy $\operatorname{det}(A^{\mathsf{T}}A)\neq 0$, thus the pseudo-inverse $(A^{\mathsf{T}}A)^{-1}$ doesn't exist, which in turn implies that the least-squares solution is not unique.