It happens that $\pi/2 - \tan 1 \approx 0.0133886$. This may just be a numerical coincidence or there may be a good reason. But in any case, $\sin$ is very flat close to $\pi/2$, so even values that are just pretty close to $\pi/2$ get translated to values very close to $1$, hence $\sin \tan 1 \approx 0.9999103$. By continuity/smoothness the fixed point must be very close to $1$, as you say.
I'll make this more quantitative in an effort to make it more "explanatory". Suppose $f(x)$ is smooth near $x=1$ and $f(1) - \pi/2 = \delta$. Then $\sin(f(1)) = \cos(\delta) = 1 - \delta^2/2 + \cdots$. Hence $\sin f(x) - x$ at $x=1$ is $-\delta^2/2 + \cdots$. If $|\delta| \approx 0$, the difference is approximately $0$, so if a fixed point exists, it should be near here, which can be made rigorous by comparing the derivative of $\sin f(x) - x$ to $-\delta^2/2 + \cdots$ at $x=1$.
In the present case with $f = \tan$, we find $\delta \approx -0.0133886$, we have $-\delta^2/2 \approx -0.0000896273$, and this is tiny compared to $(\sin \tan x - x)'(1) \approx -0.954138$, so we'll definitely get a fixed point very close to here. Indeed, the actual fixed point differs from $1$ by $0.0000939875$, which is very close to my estimate, and moreover $0.0000896273/0.954138 \approx 0.0000939354$. That is, the closeness of the fixed point to $1$ is just an amplified version of the closeness of $\tan 1$ to $\pi/2$ thanks to the sine function.
Some thoughts on $\pi/2 - \tan 1 \approx 0.0133886$: we can change the problem in at least two ways--replace $\sin$, $\pi/2$, $1$; replace $\tan$--and sometimes the above reasoning will still go through. There are a lot of potential tweaks, so I'd be satisfied by the coincidence explanation. On the other hand, it would be more interesting if was something really was going on with $\tan$ and $\pi/2$.