The hexagon with opposite sides identified is the topological torus, see here and here.
This would suggest to me that the fundamental group of the torus could be written as $\langle x,y,z|xyzx^{−1}y^{−1}z^{−1}\rangle$, but apparently this is not true (as mentioned here, for example). I don't understand why not.
Can the fundamental group of the torus be found starting from the hexagon with opposite sides identified?
The method that you used to write down a presentation using the hexagon gluing diagram is valid only under a strong hypothesis: the gluing diagram has one vertex cycle.
But the hypothesis of one vertex cycle is not satisfied by the hexagon gluing diagram, which has two vertex cycles. Writing the vertices in order around the hexagon as A, B, C, D, E, F, then A, C, E forms one vertex cycle, and B, D, F forms the second vertex cycle.
The reason for this hypothesis is that when you form the quotient surface of the gluing diagram by gluing edge pairs as indicated, the image of the edges forms the 1-skeleton of the quotient surface, and you need a unique vertex in the 1-skeleton in order for each edge to close up into a generator of the group.
You can, on the other hand, obtain a presentation even when there are two or more vertex cycles, but to do that you must first choose a maximal tree in the 1-skeleton of the quotient surface, color those edges "red", then color the corresponding edge pairs of the polygon "red", and then you simply ignore the red edges when you write the presentation (what's happening topologically is that you are taking a further quotient by collapsing the maximal tree to a point).
So, for instance, with the hexagon gluing, one can choose a maximal tree consisting of a single red edge whose corresponding edge pair in the hexagon boundary is $z,z^{-1}$. So following the formula, you ignore $z$ in the generators and your ignore $z$ and $z^{-1}$ in the relators, and you get the presentation $$\langle x, y \mid x y x^{-1} y^{-1} \rangle $$
