So S' and S are of equal age when they synchronize. Then there is 1 second of acceleration. At then end, S is now much older according to S', and the time when S synchronized his clock was a long time ago according to S', so it's no problem that S' watches him age slowly as they close in on each other.
Update: So the way this question is phrased isolates the heart of the Twin Paradox. It's not that the traveling twin ages less, it's that he ages less while seeing the at-home twin's clock run slower than his the whole time.
The paradox is resolved when you realize the at home twin's age jumps forward a lot during the turn around (acceleration). But: the acceleration is brief in BOTH reference frames...how can you account for years when each twin experienced only seconds?
That is the Andromeda Paradox: by changing your velocity, the definition of "now" at distance places changes. S' is far away and at rest w.r.t to S, so they can synchronize their clocks. Call those events $s_{\mu}'$ and $s_{\mu}$ respectively. They occur at the same time ($s_0 = s'_0$), with a large space like separation in the initial frames.
Once $S'$ is moving (fast) towards $S$, even if he accelerated only for a second in his frame and 1.07 s in $S$, the time coordinate of $s_0$ is way in his past, only because his definition of "now" at $s_i$ jumped into the future, relative to the initial conditions.
Worse: this change is reversible, just by turning his spaceship around.
To quote David Mermin:
"That no inherent meaning can be assigned to the simultaneity of distant events is the single most important lesson to be learned from relativity."