You can't get away from special relativity, which is what "unifies" electric and magnetic phenomena. The electromagnetic field really is a single field (actually a tensor, but don't worry about that), and its components mix together in the Lorentz transform in a similar way to how the x,y,z components of a usual vector field mix together in a rotation.
There are two ways to find the electric field here: You can perform a kinematic Lorentz transform to obtain the source charges/currents in the moving frame. You can also obtain the electric field component directly from the Lorentz transform of the electromagnetic field.
Lorentz Transform on Sources
An object in motion appears length contracted to observers: $L' = L_0/\gamma$, where $L_0$ is the length of the object its rest frame and $\gamma$ is the Lorentz factor associated with the velocity of the object:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
Consequently, if the object is charged, the charge density increases when in motion:
$$\rho' = \gamma \rho_0$$
The magnetic field in the rest frame arises from the magnetization $\vec M$ (magnetic dipole density) of the magnets, which is pointed up along the N-S axis. Each magnetic dipole is effectively a small current loop with current $\vec J = \rho \vec u$, where $\vec u$ is the tangential velocity around the N-S axis.
Then in the moving frame, the speed $u$ is not uniform around the loop, and charge densities at the points of the loop flowing with and against the direction of motion are different:
$$u_\pm = \frac{u \pm v}{1 \pm uv/c^2} $$
$$\rho_\pm = \gamma_\pm \rho_0 = \frac{\rho_0}{\sqrt{1 - u_\pm^2/c^2}}$$
Therefore $u_+ > u_-$, which means $\rho_+ > \rho_-$, so each current loop (magnetic dipole) acquires an electric dipole moment. Each magnetic dipole pointed up, so each electric dipole is pointed to the right. This is the source of the electric field in the moving frame.
Lorentz Transform of E and B fields
If you begin with fields $\vec E$, $\vec B$, and want the fields in an inertial frame moving at relative speed $\vec v$, the new fields are:
\begin{align}
\vec E_{\parallel}' &= \vec E_{\parallel} \\
\vec B_{\parallel}' &= \vec B_{\parallel} \\
\vec E_\perp' &= \gamma(\vec E_\perp + \vec v \times \vec B) \\
\vec B_\perp' &= \gamma(\vec B_\perp - \frac{1}{c^2} \vec v \times \vec E)
\end{align}
In our case $\vec E = 0$ and $\vec B_\parallel = 0$, so the above simplifies to
\begin{align}
\vec E' &= \gamma \vec v \times \vec B \\
\vec B' &= \gamma \vec B
\end{align}
If the $\vec B$ field is up (from S to N) and $\vec v$ is into the page, this gives an $\vec E$ field pointed to the right.
