(This is a long comment rather than a complete answer.)
As Igor Rivin points out, the mapping class group is not ${\mathbb Z}_2$. There is another ${\mathbb Z}_2$ direct summand coming from a homeomorphism that reverses the orientation on both $S^1$ and $S^2$, hence preserves the orientation of $S^1\times S^2$. This homeomorphism can be taken to have order 2 and it seems likely that $S^1\times S^2$ can be triangulated so that the homeomorphism preserves this triangulation. On the other hand, it seems somewhat unlikely that the twist homeomorphism generating the other ${\mathbb Z}_2$ can be realized by a symmetry of a triangulation. This symmetry would have finite order, and the only reason the twist along $S^2$ has order 2 is that $\pi_1SO(3)$ is ${\mathbb Z}_2$, which depends on a well-known but slightly subtle deformation of a $4\pi$ rotation of space to the identity map, the so-called belt trick. In one lower dimension a Dehn twist on a nontrivial curve on a torus, for example, has infinite order in the mapping class group, so it cannot be realized by a symmetry of a fixed triangulation of the torus. It would be a little surprising if 3 dimensions was different in this regard.