This is not quite true, because after splitting along the horizontal surface you may get interval bundles over non-orientable surfaces, and these are not products (I am assuming your 3-manifold is orientable). This holds for instance if the manifold is a circle bundle over a non-orientable 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.