This Lagrange multiplier does go to infinity to keep $\lambda \cdot \cos(x \cdot y) = 1/2$. What I mean is that this procedure of $\alpha \rightarrow 1$ is always going to give two solutions from your problem, no matter how close $\alpha$ to $1$. In particular, the following two statements are identical ( $\epsilon$ and $\zeta$ are positive and we confine $x \cdot y \in \{0, 2 \pi \}$ here):

1. find min $x^2 + y^2 $ s.t. $\sin(x \cdot y ) = 1 -\epsilon, \epsilon \rightarrow 0$
2. find min $x^2 + y^2 $ s.t. $x \cdot y = \pi/2 \pm \zeta, \zeta \rightarrow 0$

while in the second statement we have two constraint surfaces, though they may be very close. The two points found by the Lagrange multiplier are not on a simply connected path on the constraint surface, I believe that this makes this problem special.

I prefer to use NMinimize to deal with this case. It seems that this function has an option, SimulateAnnealing, that handles this discrete domain very well. You can give it a shot (the solver starts from the red point and it end at the blue point in the contour plot):

res2 = Reap@
   NMinimize[{x^2 + y^2, Sin[xy] == 1}, {x, y}, 
    EvaluationMonitor :> Sow[{x, y}], Method -> "SimulatedAnnealing"];
start = res2[[2, 1, 1]]
end = res2[[2, 1, -1]]