Instead of extending the state space model with a marginally stable state (eigenvalue of one) which is uncontrollable, one can try and see if a coordinate transformation is possible, similar to the continuous time variant discussed here.
If your system is of the form
$$
x_{k+1} = A\,x_k + B\,u_k + f \\
y_k = C\,x_k + D\,u_k + g \tag{1}
$$
with $x_k$, $u_k$ and $y_k$ the state, input and output of the system respectfully at time step $k$, $f$ and $g$ constant vectors. The following coordinate transformation could be used to make the affine system linear $z_k=x_k+\alpha$ and $v_k=u_k+\beta$ with $\alpha$ and $\beta$ constant vectors which satisfy
$$
\begin{bmatrix}
A-I & B \\ C & D
\end{bmatrix}
\begin{bmatrix}
\alpha \\ \beta
\end{bmatrix} =
\begin{bmatrix}
f \\ g
\end{bmatrix}. \tag{2}
$$
This can always be solved if the $(A-I,B,C,D)$ matrix is full rank and if this is not the case when the vector of $(f,g)$ lies in the span of the $(A-I,B,C,D)$ matrix. After this transformation the dynamics will simply be
$$
z_{k+1} = A\,z_k + B\,v_k \\
y_k = C\,z_k + D\,v_k. \tag{3}
$$
It can be noted that when applying a control technique to $(3)$, which would drive $z_k$ for example to the origin, would drive the state $x_k$ from $(1)$ to a value that is offset by $-\alpha$.
Best Answer
Just figured it out!
Wherever you see an $x$, replace it with $x + (I - A)^{-1}\,b$. In particular, for the evolution equation you obtain \begin{align*} x_{k+1} + (I - A)^{-1}\,b &= A\,(x_{k} + (I - A)^{-1}\,b) + b \\ x_{k+1} &= A\,x_{k} + (A - I)(I - A)^{-1}\,b + b \\ &= A\,x_k - b + b \\ &= A\,x_k. \end{align*}
You're welcome, past me.