Dehn twist as isometries on hyperbolic surface [I am editing the question to a most correct and precise one thanks to comments of Lor and studiosus]
Let (S,g) be a compact hyperbolic surface. On a simple closed geodesic $\gamma
$ I can realized a Dehn twist. This gives me a new hyperbolic metric on $S$.
I don't understand the following remark in Thurston's book "Three dimensional geometry and topology" middle of p. 272:

twist parameters that are the same modulo 1 result in surfaces that are isometric

The same kind of sentence seems to be in the second paragraph of section 10.7.3 of Farb and Margalit book "a primer on mapping class groups"

Note that $X_s$ [a point in Teichmüller space obtained from a given point by performing a twist of amount s along a given simple closed curve] is isometric to $X_{s+2\pi}$

 A: A Dehn twist is not an isometry of the hyperbolic surface: to convince you about this, consider any curve $\eta$ transversally intersecting $\gamma$. As you perform the Dehn twist around $\gamma$, $\eta$ has to travel along $\gamma$ for a complete turn, before crossing it. Therefore its length increases proportionally to the length of $\gamma$.
You can also have a look at this answer on MO about the failure of Mostow Rigidity in dim 2.
[Additional comments]
Maybe looking at the genus 1 case is easier to visualize since the geometry is Euclidean. The Teichmüller space for the torus can be identified with the upper half plane $\mathcal{H}\subseteq\mathbb{R}^2$. A point $z\in \mathcal{H}$ determines a flat (=Euclidean) structure on the torus in this way: just take the Euclidean quadrilateral with vertices $\{0,1,1+z,z\}$ and identify opposite edges by a translation. The standard flat unit torus $T$ corresponds to $z=i$. Let $\alpha,\beta$ denote the curves on $T$ corresponding to the segments $[0,1]$ and $[0,i]$ in the universal cover $\widetilde{T}$. A positive Dehn twist along $\alpha$ stretches the curve $\beta$; you can visualize this on the universal cover by sliding the segment $[i,i+1]$ to the right till it becomes the segment $[i+1,i+2]$. In other words you have deformed the unit square with vertices $\{0,1,1+i,i\}$ to the quadrilateral $\{0,1,1+2,i+1\}$. In  the Teichmüller space $\mathcal{H}$ we have gone from the point $i$ o the point $i+1$. Now we observe that
1) the transformation we have performed is not an isometry: the length of $\beta$ in the first torus is the same as the length of the segment $[0,i]$ i.e. 1, but in the second torus its length is the same as that of the segment $[0,i+1]$ i.e. $\sqrt{2}$. But an isometry should preserve the length of geodesics. Or you can check the angle between the two geodesics.
2) the two tori we obtained are nevertheless isometric: to see this just cut the square $\{0,1,1+i,i\}$ along the diagonal $[0,i+1]$ and identify the vertical segments (e.g. by a translation); you now have a fundamental domain for the second torus.
Possible source of confusion: the two tori are isometric, hence they give the same point in the Moduli space, but have different markings, so they are represented by different points in the Teichmüller space; the action of the mapping class group sends the $i$-torus to the $(i+1)$-torus via the Dehn twist described above. The phenomenon is indeed the same for higher genus surfaces.
