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 $\epsilon < 1$?



Is it impossible for determinants of these matrices to both be negative?
This is a partial answer, now with added material.


Assume that $n$ is a multiple of 3, the other two cases should be similar. I'll denote $C\_1\mapsto a$ and $C\_2\mapsto b$, such that for, e.g., $n=6$,
$$\tag{1}\label{1}
A = \begin{pmatrix}
a\_1 & 1 & 0 & 0 & 0 & 0 \\
a\_2 & 0 & 1 & 0 & 0 & 0 \\
a\_3 & 0 & 0 & 1 & 0 & 0 \\
a\_4 & 0 & 0 & 0 & 1 & 0 \\
a\_5 & 0 & 0 & 0 & 0 & 1 \\
a\_n & 0 & 0 & 0 & 0 & 0 
\end{pmatrix},\quad
A^{-1} = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 1/a\_n \\
1 & 0 & 0 & 0 & 0 & -a\_1/a\_n \\
0 & 1 & 0 & 0 & 0 & -a\_2/a\_n \\
0 & 0 & 1 & 0 & 0 & -a\_3/a\_n \\
0 & 0 & 0 & 1 & 0 & -a\_4/a\_n \\
0 & 0 & 0 & 0 & 1 & -a\_5/a\_n
\end{pmatrix}.
$$
Note that $\det A = (-1)^{n-1}a\_n$. The inverse of $A$ is easily calculated, see \eqref{1}, such that the determinant of $AB+A+I$ (and also of the other case, exchange $a\_i$ and $b\_i$) can be expressed via a Schur complement of the matrix $A^{-1}(AB+A+I) = B+I+A^{-1}$:
We take the Schur complement w.r.t. the first and last row/column and get
$$\tag{2}\label{2}
\det(AB+A+I)=\det
\begin{pmatrix}
\alpha\_1-\alpha\_3 & \beta\_1-\beta\_2\\
\alpha\_2-\alpha\_3 & \beta\_1-\beta\_3
\end{pmatrix}
=\det
\begin{pmatrix}
\alpha\_1 &1& \beta\_2\\
\alpha\_2 &1& \beta\_3\\
\alpha\_3 &1& \beta\_1
\end{pmatrix},
$$
where in the last step we used basic determinant rules.
The $\alpha\_j$ and $\beta\_j$, with $j\in\{1,2,3\}$, are simply given by
$$\tag{3}\label{3}
\alpha\_j = -\delta\_{3,j} + \sum\_{k=0}^{n/3-1} a\_{3k+j},
\qquad
\beta\_j = -\delta\_{3,j} + \sum\_{k=0}^{n/3-1} b\_{3k+j},
$$
with Kronecker's $\delta$. I’ve renamed OPs $\beta\mapsto\nu$.


For the eigenvalue assumption $|\lambda\_i|, |\nu\_i| <1$ it should be sufficient to consider the characteristic polynomials
\begin{align}\tag{4a}\label{4a}
P\_a(\lambda)&=\det(A^2 B - \lambda I),\\
P\_b(\nu )&=\det(A B^2 - \nu I),\tag{4b}\label{4b}
\end{align}
which can be calculated in a similar fashion: now we use $A^{-1}(A^2 B - \lambda I) = A B - \lambda A^{-1}$ and build the Schur complement w.r.t. the first, second and last row/column.


If the eigenvalues fulfill $|\lambda\_i|<1$, the (real) zeroes of $P\_a(\lambda)$ are between $\lambda=\pm 1$. At this point, I am not sure how to handle the complex zeroes (but see edit below). However, as $P\_a(\lambda) \sim (-\lambda)^n$ for large $|\lambda|$, it should fulfill $P(-1)>0$ and $(-1)^{n}P(1)>0$. Let's evaluate $P\_a(1)$ as one example,
$$\tag{5}\label{5}
P\_a(1)=\det(A^2B-I)=
\det
\begin{pmatrix}
\alpha\_1 & \alpha\_3 & \beta\_2\\
\alpha\_2 & \alpha\_1 & \beta\_3\\
\alpha\_3 & \alpha\_2 & \beta\_1
\end{pmatrix}.
$$
Note the similarity of \eqref{2} and \eqref{5}, and that \eqref{5} can be generalised to $P\_a(\lambda)$ using polynomials $\alpha\_j(\lambda),\beta\_j(\lambda)$.
So, the problem for arbitrary $n$ can be reduced to a discussion of the sign of the determinants of related $3\times3$ matrices.


Disclaimer: there might be sign errors due to even/odd $n$ and row/column permutations, please check.


**Edit 12.08.22, 09:00 CEST:**


From now on, we only consider even $n$, such that $n\mod 6 \equiv 0$, to get rid of the $(-1)^n$ terms.


As noted in my comments, $P\_a(1)>0$ and $P\_a(-1)>0$ are necessary conditions for $|\lambda\_i|<1$, because complex $\lambda\_i$ appear in complex conjugate pairs $\lambda\_{i'}=\lambda\_i^\*$, such that
$$\tag{6}\label{6}
|\lambda\_i|<1 \Rightarrow (\lambda\_i \pm 1)(\lambda\_i^\* \pm 1)>0,
$$
and all factors in
$$\tag{7}\label{7}
P\_a(\lambda) = \prod\_{i=1}^{n/2} (\lambda\_i - \lambda)(\lambda\_{i'} - \lambda)
$$
are positive for $\lambda=\pm 1$. Here, we grouped the real eigenvalues in arbitrary pairs $(i,i')$.


As shown above, $P\_{a,b}(1)$ have the simple representation \eqref{5}.
Hence, we consider the matrix
$$\tag{8}\label{8}
D(\alpha,\gamma,\beta)=\begin{pmatrix}
\alpha\_1 &\gamma\_3& \beta\_2\\
\alpha\_2 &\gamma\_1& \beta\_3\\
\alpha\_3 &\gamma\_2& \beta\_1
\end{pmatrix}
$$
with 3d vectors $\alpha,\beta,\gamma$, and formulate a geometric version of the problem. Define $\delta=(1,1,1)^T$, then the OPs conjecture holds, if
$$\tag{9}\label{9}
D(\alpha,\alpha,\beta)>0 \land D(\alpha,\beta,\beta)>0
\Rightarrow
D(\alpha,\delta,\beta)>0 \lor D(\beta,\delta,\alpha)>0.
$$
Note (a) that the determinant in 3D is known as triple product,
$$\tag{10}\label{10}
\det(a,b,c) = a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b),
$$ and (b) that the cyclic index permutations of in \eqref{2}, \eqref{5} and \eqref{8} are rotations by $120^\circ$ around $\delta$. I guess that the OP question can now be answered through a discussion of the (rotated) directions of $\alpha$, $\beta$ and $\delta$ in 3D.


