As a matter of preference, I will use $\mathbf{Q}_{o}$ and $\delta \mathbf{Q}$ for the quasi-static and fluctuating terms.
Next, recall from vector calculus that the following holds:
$$
\mathbf{A} \times \left( \mathbf{B} \times \mathbf{C} \right) = \left( \mathbf{A} \cdot \mathbf{C} \right) \mathbf{B} - \left( \mathbf{A} \cdot \mathbf{B} \right) \mathbf{C} \tag{1}
$$
Next we use the expression for $\delta \mathbf{v}$ in the convective electric field term to get:
$$
\begin{align}
\delta \mathbf{E} & = - \frac{ 1 }{ c } \ \delta \mathbf{v} \times \mathbf{B}_{o} \tag{2a} \\
& = - \frac{ i }{ \rho_{o} \ \omega \ c } \left[ \left( \delta \mathbf{j} \times \mathbf{B}_{o} \right) \times \mathbf{B}_{o} \right] \tag{2b} \\
& = \frac{ i \ B_{o}^{2} }{ \rho_{o} \ \omega \ c^{2} } \left[ \frac{ \mathbf{B}_{o} \times \left( \delta \mathbf{j} \times \mathbf{B}_{o} \right) }{ B_{o}^{2} } \right] \tag{2c} \\
& = \frac{ 4 \ \pi \ i }{ \omega } \left( \frac{ V_{A} }{ c \ B_{o} } \right)^{2} \left[ B_{o}^{2} \ \delta \mathbf{j} - \left( \delta \mathbf{j} \cdot \mathbf{B}_{o} \right) \mathbf{B}_{o} \right] \tag{2d}
\end{align}
$$
The next thing to notice is that Equation 2a tells us that $\delta \mathbf{E}$ is orthogonal to $\mathbf{B}_{o}$, therefore the last term in Equation 2d must be zero. This is another way of saying there are no field-aligned current perturbations. Then we can rewrite Equation 2d in terms of the current to get:
$$
\delta \mathbf{j} = \frac{ \omega }{ 4 \ \pi \ i } \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} \tag{3}
$$
Finally, Ampere's law goes to:
$$
\begin{align}
i \mathbf{k} \times \delta \mathbf{B} & = \frac{ 4 \ \pi }{ c } \delta \mathbf{j} - \frac{ i \ \omega }{ c } \delta \mathbf{E} \tag{4a} \\
\frac{ i \ c }{ \omega } \mathbf{k} \times \left( \mathbf{k} \times \delta \mathbf{E} \right) & = \frac{ 4 \ \pi }{ c } \left[ \frac{ \omega }{ 4 \ \pi \ i } \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} \right] - \frac{ i \ \omega }{ c } \delta \mathbf{E} \tag{4b} \\
\mathbf{k} \times \left( \mathbf{k} \times \delta \mathbf{E} \right) & = - \left( \frac{ \omega }{ c } \right)^{2} \left( \frac{ c }{ V_{A} } \right)^{2} \delta \mathbf{E} - \left( \frac{ \omega }{ c } \right)^{2} \delta \mathbf{E} \tag{4c} \\
& = - \left( \frac{ \omega }{ c } \right)^{2} \left[ 1 + \left( \frac{ c }{ V_{A} } \right)^{2} \right] \delta \mathbf{E} \tag{4d} \\
& = - \omega^{2} \left[ \frac{ 1 }{ c^{2} } + \frac{ 1 }{ V_{A}^{2} } \right] \delta \mathbf{E} \tag{4e} \\
& = - \left( \frac{ \omega }{ V_{A} } \right)^{2} \left[ 1 + \left( \frac{ V_{A} }{ c } \right)^{2} \right] \delta \mathbf{E} \tag{4f}
\end{align}
$$
where Equation 4f is the same as that which you seek.
The only part that may not be immediately obvious here is that $\delta \mathbf{j}$ is orthogonal to $\mathbf{B}_{o}$. Generally if the only contribution to the electric field is from the convective term, then there can be no field-aligned currents. If you allow for other terms in the generalized Ohm's law, then you can get field-aligned currents, which are equivalent to a specific form of Alfven wave called kinetic or shear Alfven waves, depending on the limits/boundary conditions of interest.
