The Liouville von Neumann equation is
$$\frac{\partial}{ \partial t} \rho(t)=-i [H, \rho(t)]$$
Now, the hamiltonian is
$$ H= H_0+H_1$$
and I want to transform it into the interaction Picture, so
$$ \rho_I=\exp(i H_0 t) \rho \exp(-i H_0 t)$$.
Then
$$\frac{\partial}{\partial t} \rho_I= i H_0 \rho_I + \exp(i H_0 t) (\frac{\partial }{\partial t} \rho) \exp(-i H_0 t) - i \rho_I H_0$$
Writing it as a commutator and plugging in the Liouville von Neumann equation, this can be written as
\begin{align} \frac{\partial}{\partial t} \rho_I &= i [H_0 ,\rho_I] + \exp(-i H_0 t) (-i [H_0+H_I, \rho(t)])\exp(i H_0 t) \\ &=i[H_0, \rho_I]-i [H_0, \rho_I]-i \exp(i H_0)[H_1, \rho_I] \exp(-i H_0 t) \\ &=-i \exp(i H_0 t) [H_1, \rho] \exp(-i H_0 t) \end{align} Now, if $[H_o, H_1]=0$, this can be cast in the form that I know
$$ \frac{\partial}{\partial t} \rho_I=-i [H_1, \rho_I] \quad (1)$$
But if they don't commute, this is not possible. So why is the Liouville von Neumann equation normally written in the form of equation (1), even though it might be possible that $H_0$ and $H_1$ don't commute? Am I making a mistake in my derivation? Is there a smarter way to derive this equation?