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Igor Belegradek
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As mentioned in the comments, if the limit is a circle, then by the Yamaguchi fibration theorem, $M_i$ fibers over the circle, and hence it is a torus (or Klein bottle in the non-orientable case).

If the limit is a segment, then $M_i$ is $S^2$ for all large $i$. One (somewhat heavy handed) way to see this is to apply Corollary 0.4 of Shioya-Yamaguchi's paper. Indeed, the product of $M_i$ and a unit circle, collapses to the cylinder. Hence in Corollary 0.4 we have $g=0$ and $k=2$. Hence for large $i$ the fundamental group of $M_i\times S^1$ is a free product of $\mathbb Z$ and finitely many finite cyclic groups. Such a group cannot have a center unless all the finite cyclic groups are trivial. The circle factor of $M_i\times S^1$ is central in the fundamental group, so $\pi_1(M_i\times S^1)=\mathbb Z$. Hence $M_i$ is a sphere (for large $i$).

This argument uses orientability of $M_i$ because Shioya-Yamaguchi only deal with orientable manifolds.

Igor Belegradek
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  • 176