Is it impossible for determinants of these matrices to both be negative? Suppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$).
Thus we have $B = A+ \xi e_{1}^T$, where $\xi$ is a $n \times 1$ column vector and $e_{1}^{T} = [1, 0,\ldots,0]$
$\textbf{Assumptions}:$
Let $\lambda_{i}, i=1, \ldots, n$ denote the eigenvalues  of $AB^2$. Suppose we have the condition that $|\lambda_{i}|<1 \, \forall i$.
Let $\beta_{i}, i=1, \ldots, n$ denote the eigenvalues  of $A^2B$. Suppose we have the condition that $|\beta_{i}|<1 \, \forall i$.
$\textbf{Claim}:$
Then I have an intuition that $\textbf{ $\det(AB+A+I) < 0$ and $\det(BA+B+I)<0$ is not possible.}$
That is both of the determinants cannot be negative. I am not sure how to prove it?
$\textit{Some thoughts:}$
$(a) \textbf{Using $A = B+ \xi e_{1}^T$, we have }$
\begin{align}
AB+A+I &= A^2 + A + I + A \xi e_{1}^{T},\\
BA+B+I &= A^2 + A + I + \xi e_{1}^{T}(I+A)
\end{align}
Since the absolute values of the eigenvalues of $AB^2,$ and $A^2B$ are less than one, that means we have $|\det(AB^2)| < 1$ and $|\det(A^2B)|<1$.
\begin{align}
A^2B &= A^3 + A^2 \xi e_{1}^{T},\\
AB^2 &= A^3 + (\xi e_{1}^{T})^2 + 2A^{2} \xi e_{1}^{T}
\end{align}
$\textbf{(b)}$ If we proceed via the method of contradiction. if $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$, then some of the eigenvalues of $AB^2$ or $A^2B$ will be greater than one in absolute value. THis would then violate the assumption of $|\lambda_{i}|<1 \forall i$,, $|\sigma_{i}|<1, \forall i$ .
$\textbf{(c)}$ Another thought is to use the perturbation argument: Fix $A, B$. Define $B(\epsilon):= A + \epsilon (B-A)$. For $\epsilon = 0$, we get $\det(A^2+A+I) \geq 0$ and hence the statement holds. For $\epsilon=1$, we have $B(1) = B$. If the statement fails in this case then there should be a minimal $\epsilon$ for which the statement is false. There might be a contradiction for $\epsilon < 1$?
 A: OK, second try, equation references are pointing to my other answer:
Using the definitions (3) for $\alpha_j$ and $\beta_j$, we can proof the OPs conjecture in the following way: As the determinants
$$
\begin{align}
D_a = \det(AB+A+I) \tag{11a}\label{11a}\\
D_b = \det(BA+B+I) \tag{11b}\label{11b}
\end{align}
$$
are continuous functions of the $a_i$ and $b_i$ (and $\alpha_j$ and $\beta_j$), and both shall be negative, w.l.o.g. we can assume that, say, either $D_a=0$ crosses zero while the other is still positive, $D_b>0$, or, both are zero simultaneously. This case is handled first.
case $D_a=D_b=0$
First we show that $D_a$ and $D_b$ cannot simultaneously vanish under the OPs eigenvalue condition $|\lambda_i|<1$.
As can be seen from (2), w.l.o.g. the condition $D_a = D_b = 0$ is equivalent to $\alpha_1=\alpha_2=\alpha_3$.
(Note that this condition can be solved for, e.g., $a_1$ and $a_2$.)
Inserting this solution into (5), we immediately see that the first two columns of (5) become linearly dependent, such that the characteristic polynomial (4a) reads $P_a(1)=0$. Therefore, we have identified one eigenvalue $\lambda_1=1$, contradicting the assumption that all eigenvalues $|\lambda_i|<1$. The other possible case $\beta_1=\beta_2=\beta_3$ would give $P_b(1)=0$.
case $D_a=0 \land D_b>0$
Now we turn to the case $D_a=0 \land D_b>0$: Then, from (2) we see that
$$\tag{12}\label{12}
\alpha_1=\beta_2 \land \alpha_2=\beta_3 \land \alpha_3=\beta_1
$$
must hold. Inserting this into (2) for the determinant $D_b$, we get the inequality
$$
\begin{align}
D_b &= \det
\begin{pmatrix}
\beta_1 &1& \alpha_2\\
\beta_2 &1& \alpha_3\\
\beta_3 &1& \alpha_1
\end{pmatrix}
=
\det\begin{pmatrix}
\alpha_3 &1& \alpha_2\\
\alpha_1 &1& \alpha_3\\
\alpha_2 &1& \alpha_1
\end{pmatrix} \tag{13a}\label{13a} \\
\\
&= -\frac{(\alpha_1-\alpha_2)^2 + (\alpha_2-\alpha_3)^2 + (\alpha_3-\alpha_1)^2}{2} < 0 \tag{13b}\label{13b}
\end{align}
$$
which contradicts the assumption. Therefore, the only continuous path to negative $D_a$ and $D_b$ goes through $D_a=D_b=0$, at which point $\lambda_1=1$.
