Why is the half-torus rigid?
I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:
half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or Ефимов, УМН, 1948, т.3, вып.2, стр. 135)
Any second order rigid surface does not admit analytic deformation (i.e., the deformation $h_t(u,v)$ which is analytic on $t$).
For the surfaces of revolution, the assumption of analyticity can be removed.
Below is the best part of my original post. It contains an idea which was not used by Rembs.
Let $h(u,v)$ be a small perturbation of the standard embedding; $u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$ and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$. I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$. Indeed since $\gamma_0$ lies on convex part, the Gauss curvature at the points of $\gamma_0$ has to be nonnegative. On the other hand $\gamma_0$ bounds a flat disc in $\partial K$; therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$. If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$. I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$. In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.