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The rider expends extra energy on the downstroke to flex the frame, and then gets it back on the upstroke, so the net energy loss is zero.
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Yes, exactly as you say, your legs get the energy back. And when they do, of course that energy doesn't move the bike forward. I don't think that's what we're trying to call zero net energy loss.
Put a balloon on a scale, and then put two five pound weights on the balloon. What's the scale read? Ten pounds, or close enough. Now remove one of the weights. Does the scale read seven and a half pounds? And does the expanding balloon raise the remaining weight only as high as if it indeed weighed seven and a half?
That is roughly what you're proposing, if you say released strain energy from the frame drives the bike.
When does the frame un-flex? When you reduce the pressure on the pedals. And does the resistance at the rear wheel change, then? Nope, stays the same. So, big push at one end of the frame's flex, little push at the other end -- where is the difference in push going to do some work?
It's important to distinguish frame flex induced by drive torque from that induced by other forces, but thinking about the paths your feet take as they go around, and the sideways BB sway summed with the kinking of the stays, is needless distraction. All those things happen when you ride, but they do not affect the way the system behaves. It is perfectly accurate to regard the system of frame flex induced by drive torque as two forces squeezing a spring, or weights on a balloon. And it's painfully obvious that such a system cannot behave in violation of simple physics: the lesser force cannot overcome the greater.