It seems that you are using spatial accelerations for $\dot{v}_G$. So your equations of motion at the center of mass G are correct
$$ \pmatrix{f \\ \tau_G} = \left[ \matrix{ m & \\ & J_G} \right] \pmatrix{\dot{v}_G \\ \dot{\omega}} + \pmatrix{m \omega \times v_G \\ \omega \times J_G \omega} $$
Notice I added a subscript on the MMOI at it is expressed about G and removed it from $\omega$ since it is shared by the entire rigid body.
The first parts of the equation are easily transformed to a different point B with $$\pmatrix{f \\ \tau_B} = \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right] \pmatrix{f \\ \tau_G}$$ and $$\pmatrix{\dot{v}_G \\ \dot{\omega}} = \left[ \matrix{ 1 & -\rho_G \times \\ 0 & 1} \right] \pmatrix{\dot{v}_G \\ \dot{\omega}}$$
Your question is, how do you transform the bias forces (the last part of the equations)?
Note that $$\begin{pmatrix}m\omega\times v_{G}\\ \omega\times J_{G}\omega \end{pmatrix} = \begin{bmatrix}\omega\times\\ v_{G}\times & \omega\times \end{bmatrix}\begin{bmatrix}m\\ & J_{G} \end{bmatrix}\begin{pmatrix}v_{G}\\ \omega \end{pmatrix}$$
So the equations at B are
$$ \pmatrix{f \\ \tau_B} = \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right]\left[ \matrix{ m & \\ & J_G} \right] \left[ \matrix{ 1 & -\rho_G \times \\ 0 & 1} \right] \pmatrix{\dot{v}_G \\ \dot{\omega}} +\\ + \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right]\begin{bmatrix}\omega\times\\ v_{G}\times & \omega\times \end{bmatrix}\begin{bmatrix}m\\ & J_{G} \end{bmatrix}\begin{bmatrix}1 & -\rho_{G}\times\\ 0 & 1 \end{bmatrix}\begin{pmatrix}v_{B}\\ \omega \end{pmatrix} $$$$\begin{align} \pmatrix{f \\ \tau_B} & = \underbrace{ \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right]\left[ \matrix{ m & \\ & J_G} \right] \left[ \matrix{ 1 & -\rho_G \times \\ 0 & 1} \right] }_\mbox{6×6 spatial inertia at B} \pmatrix{\dot{v}_G \\ \dot{\omega}} +\\ & + \left[ \matrix{ 1 & 0 \\ \rho_G \times & 1} \right] \begin{bmatrix}\omega\times\\v_{G}\times & \omega\times\end{bmatrix} \begin{bmatrix}m\\ & J_{G}\end{bmatrix} \begin{bmatrix}1 & -\rho_{G}\times\\0 & 1\end{bmatrix} \begin{pmatrix}v_{B}\\\omega\end{pmatrix} \end{align} $$
In the end you are trying to show that the last part is $$ +\begin{bmatrix} \omega\times\\v_{B}\times & \omega\times\end{bmatrix} \begin{bmatrix}1 & 0\\\rho_{G}\times & 1\end{bmatrix} \begin{bmatrix}m\\ & J_{G}\end{bmatrix} \begin{bmatrix}1 & -\rho_{G}\times\\0 & 1\end{bmatrix} \begin{pmatrix}v_{B}\\\omega\end{pmatrix}$$
I can show this with
$$\begin{aligned} \begin{bmatrix}\omega\times\\ v_{B}\times & \omega\times \end{bmatrix}\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}&=\\\begin{bmatrix}\omega\times\\ \left(v_{G}+\rho_{G}\times\omega\right)\times & \omega\times \end{bmatrix}\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}&=\begin{bmatrix}\omega\times\\ \left(v_{G}+\rho_{G}\times\omega\right)+\omega\times\rho_{G}\times & \omega\times \end{bmatrix}\\\begin{bmatrix}\omega\times\\ v_{G}\times+\left(\rho_{G}\times\omega\right)\times+\omega\times\left(\rho_{G}\times\right) & \omega\times \end{bmatrix}&=\begin{bmatrix}\omega\times\\ v_{G}\times+\left(\rho_{G}\times\omega\times-\omega\times\rho_{G}\times\right)+\omega\times\rho_{G}\times & \omega\times \end{bmatrix}\\\begin{bmatrix}\omega\times\\ v_{G}\times+\rho_{G}\times\omega\times+ & \omega\times \end{bmatrix}&=\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}\begin{bmatrix}\omega\times\\ v_{G}\times & \omega\times \end{bmatrix} \end{aligned}$$$$\begin{aligned} \begin{bmatrix}\omega\times\\v_{B}\times & \omega\times\end{bmatrix}\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}&=\\\begin{bmatrix}\omega\times\\ \left(v_{G}+\rho_{G}\times\omega\right)\times & \omega\times \end{bmatrix}\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}&=\begin{bmatrix}\omega\times\\ \left(v_{G}+\rho_{G}\times\omega\right)+\omega\times\rho_{G}\times & \omega\times \end{bmatrix}\\\begin{bmatrix}\omega\times\\ v_{G}\times+\left(\rho_{G}\times\omega\right)\times+\omega\times\left(\rho_{G}\times\right) & \omega\times \end{bmatrix}&=\begin{bmatrix}\omega\times\\ v_{G}\times+\left(\rho_{G}\times\omega\times-\omega\times\rho_{G}\times\right)+\omega\times\rho_{G}\times & \omega\times \end{bmatrix}\\\begin{bmatrix}\omega\times\\ v_{G}\times+\rho_{G}\times\omega\times+ & \omega\times \end{bmatrix}&=\begin{bmatrix}1 & 0\\ \rho_{G}\times & 1 \end{bmatrix}\begin{bmatrix}\omega\times\\ v_{G}\times & \omega\times \end{bmatrix} \end{aligned}$$
Use the vector identity $\left(\rho_{G}\times\omega\right)\times = \rho_{G}\times\omega\times-\omega\times\rho_{G}\times$
So, in the end, the equations of motion are
$$ \begin{aligned} & \mbox{at G} & & \mbox{at B} \\ {\bf f}_G & = {\rm J}_G \dot{{\bf v}}_G + {\bf v}_G \times {\rm J}_G {\bf v}_G & {\bf f}_B & = {\rm J}_B \dot{{\bf v}}_B + {\bf v}_B \times {\rm J}_B {\bf v}_B \\ {\bf f}_G & = \pmatrix{f \\ \tau_G} & {\bf f}_B & = \pmatrix{f \\ \tau_B} \\ {\bf v}_G & = \pmatrix{v_G \\ \omega} & {\bf v}_B & = \pmatrix{v_B \\ \omega} \\ \dot{{\bf v}}_G & = \pmatrix{\dot{v}_G \\ \dot{\omega}} & \dot{{\bf v}}_B & = \pmatrix{\dot{v}_B \\ \dot{\omega}} \\ {\rm J}_G & = \begin{bmatrix}m\\ & J_{G}\end{bmatrix} & {\rm J}_B & = \begin{bmatrix}m & -m \rho_G \times\\m \rho_G \times & J_{G}-m \rho_G \times \rho_G \times\end{bmatrix} \\ {\bf v}_G \times & = \begin{bmatrix}\omega\times\\v_{G}\times & \omega\times\end{bmatrix} &{\bf v}_B \times & = \begin{bmatrix}\omega\times\\v_{B}\times & \omega\times\end{bmatrix} \end{aligned}$$
NOTE:
It is worth in to use material accelerations to derive the equation of motion like with my answer here and then transform back to spatial accelerations (with $a_B = \dot{v}_B + \omega \times v_B$) to simplify the terms.
Also, The term $J_{B}=J_{G}-m \rho_G \times \rho_G \times$ is the vector form of the parallel axis theorem. All throughout I am using $\times$ as the 3×3 cross product operator $$r \times = \pmatrix{x\\y\\z} \times = \left[ \matrix{0&-z&y\\z&0&-x\\-y&x&0} \right]$$