I want to prove the following assumption: Let $g,h$ be natural numbers with $g > h$ and let $S_g$ be the closed, orientable surface of genus $g$. Then, there is no (smooth) map $f: S_g \to S_h \times (-1,1)$, embedding $S_g$ as an incompressible subsurface.
The problem here is that it's not sufficient to look at the fundamental group alone. For any $h$ there exists some $g > h$ and a covering map from $S_g$ to $S_h$. This will induce a smooth map $f: S_g \to S_h \times (-1,1)$ which will provide an injection on fundamental groups. Clearly, however, this $f$ is not embedding.
Similarily, one can easily also construct an embedding on $S_g$ into $S_h \times (-1,1)$, but such a map will not be incompressible.
I tried to argue by contradiction, and came up with the following idea: Suppose there existed an incompressible embedding $f: S_g \to S_h \times (-1,1)$. This would induce a map $f_*: H_1(S_g) \to H_1(S_h)$ which must have non-trivial kernel, since $g > h$. As such, one can find an essential simple closed curve $\gamma \subset S_g$ such that it's embedded counterpart $f(\gamma)$ is null-homologous in $S_g \times (-1,1)$. Now, somehow, the fact that $f$ is an embedding must imply that $f(\gamma)$ is already null-homotopic. How can we show this ?
Here is a sketch (let me know if you want to see more details on any of these items) of a direct proof which does not require hard classification results which appear in the links, it inly requires differential topology which you can find the the book of Guillemin and Pollack:
Each nonempty connected closed subsurface $S\subset M= int(S_h\times [-1,1])$ separates the manifold $M$ in two components.
If $S$ does not separate the boundary components of $N=S_h\times [-1,1]$ then $S$ represents the trivial homology class of $N$, which implies that the inclusion map $S\to N$ cannot be homotopic to a covering map $S\to S_h$, in particular, $S$ cannot be incompressible.
Thus, $S$ separates the boundary components. Then the algebraic intersection number between $S$ and any arc $a$ of the form $x\times [-1,1]$ is $\pm 1$ (depending on the orientation). Now, compute the same algebraic intersection number for $a$ and a degree $d$ covering map $S_g\to S_h$. You get $\pm d$. Therefore, since the algebraic intersection number is homotopy invariant for maps of closed surfaces in to $N$, it follows that $d= \pm 1$.
Now, use Euler characteristic to check that for each covering map $S_g\to S_h$ the degree is $\pm 1$ iff $g=h$ and the covering is trivial.
To conclude: $g=h$ for each incompressible surface.