Since $\det(A) \neq 0$, $A$ is nonsingular and all diagonals of $A$ must equal one. Next, we know the relationship between the determinant and trace is \begin{align*} \det(A) &= \exp(tr(\ln(A))) \\ &= \exp(0) \\ &= \exp(\sum_i \ln(\lambda_i)) \\ \end{align*}. Step 2 holds because all diagonals of $A$ are one, so their log is zero, and step 3 because natural logs are analytic. This suggests $\sum_i \ln(\lambda_i) = 0$, or, $\prod_i \lambda_i = 1$.
Then, because all diagonals of $A$ are one, $tr(A) = \sum_i \lambda_i = n$ must be true. Hence, the AM-GM inequality states $ \left[ \prod_{i} \lambda_i \right]^{\frac{1}{n}} \leq \frac{1}{n} \sum_{i} \lambda_{i}$, with equality holding only when $\lambda_i = \lambda_j$ $\forall i,j$. Thus, $ 1^{\frac{1}{n}} = \frac{1}{n} \cdot n$, and all eigenvalues of $A$ must be equal. Since $\prod_i \lambda_i = 1$, all eigenvalues must be one.