This is not quite true, because after splitting along the horizontal surface you may get interval bundles over nonorientable surfaces, and these are not products (I am assuming your 3manifold is orientable). This holds for instance if the manifold is a circle bundle over a nonorientable surface (with Euler number zero, so that it has indeed a horizontal surface).
If I am not missing something, the following facts should be equivalent on an orientable Seifert fibered space $M$ without boundary:

The fibration has a horizontal surface

The fibration has Euler number zero

The manifold $M$ either fibers over $S^1$ with periodic monodromy, or it fibers over the interval orbifold, in such a way that its natural double cover fibers over $S^1$ with periodic monodromy
 The manifold $M$ has some cover (of any finite degree) that fibers over $S^1$ with periodic monodromy
 The manifold $M$ is finitely covered by $S\times S^1$ for some closed orientable surface $S$.
 The manifold $M$ has a geometry of type $S^2 \times \mathbb R$, $\mathbb R^3$, or $\mathbb H^2 \times \mathbb R$
There are few cases where the fibration is not unique, but "having a horizontal surface" or "having Euler number zero" are properties that hold for some fibration if and only if they hold for any fibration.