= Solution
Write horizontal components with a subscript $h$. The steady <ideal magnetohydrodynamic induction equation> gives $\nabla\times(\mathbf u\times\mathbf B)=0$. Since $\partial_z=0$, its horizontal components imply
$$
\partial_x(\mathbf u\times\mathbf B)_z=\partial_y(\mathbf u\times\mathbf B)_z=0.
$$
Thus $(\mathbf u\times\mathbf B)_z=C$ is a spatial constant. A vanishing axial <electric field>, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and $\mathbf B_h\ne0$, define the <magnetohydrodynamic mass loading> $k$ by $\rho\mathbf u_h=k\mathbf B_h$, and put $U=u_z-kB_z/\rho$. Then
$$
\boxed{\mathbf u=\frac{k\mathbf B}{\rho}+U\mathbf e_z}.
$$
The steady <continuity equation> and the <solenoidal magnetic-field constraint> give
$$
0=\nabla\cdot(\rho\mathbf u)=\nabla\cdot(k\mathbf B+\rho U\mathbf e_z)=\mathbf B\cdot\nabla k.
$$
Also $\mathbf u\times\mathbf B=U\mathbf e_z\times\mathbf B$. Using the vector identity for the curl of a cross product, together with $\partial_z\mathbf B=0$ and $\nabla\cdot\mathbf B=0$, its curl is $\mathbf e_z\mathbf B\cdot\nabla U$. Hence
$$
\boxed{\mathbf B\cdot\nabla k=\mathbf B\cdot\nabla U=0}.
$$
These are <steady planar ideal-MHD field-line invariants>: both quantities are constant along a <magnetic field line>.
The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho,p,s$, $\Phi=0$, $\mathbf u=u_0\mathbf e_x$ and $\mathbf B=B_0\mathbf e_y$, with $u_0B_0\ne0$, satisfy every displayed <ideal magnetohydrodynamics> equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0B_0$ is nonzero, so no $k,U$ give the stated decomposition. \b[The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample]. The subsequent calculations use that branch.
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