The interpolation overshoots the next point and reverses direction.
ParametricPlot[{line[[1]][tt], line[[2]][tt]}, {tt, 0.2, 0.3},
Epilog -> {Point[points[[All, 1 ;; 2]]]}]
You can reduce the interpolation order to 1 or use a centripetal parametrization parametrizeCurve from J.M.'s answer.
parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
tvals = parametrizeCurve[points];
line = Interpolation[Transpose[{tvals, #}]] & /@ Transpose@points;
s[t_?NumericQ] :=
NIntegrate[Norm[{line[[1]][tt], line[[2]][tt]}],
Evaluate @ DeleteDuplicates @
Flatten[{tt, 0, Select[tvals, 0 < # < t &], t}]];
ParametricPlot[Evaluate@{s[ss], line[[3]][ss]}, {ss, 0, 1},
PlotRange -> {{0, All}, {0, All}}, AspectRatio -> 1/2,
Epilog -> {Point@Table[{s[ss], line[[3]][ss]}, {ss, 0, 1, 0.1}]}]
Including the interpolation grid points tvals in NIntegrate speeds up the integration. For a really fast implementation use
s = NDSolveValue[{ss'[t] == Norm[{line[[1]][t], line[[2]][t]}], ss[0] == 0}, ss, {t, 0, 1}]
which constructs an InterpolatingFunction for the arc length.