Yes, any such function is analytic.

Assume contrary, let $f$ be such a function.
Note that if it is analytic at two points then it has to be analytic everywhere between. So by taking restriction, we may assume that the function is not analytic in any subinterval.

We can assume that $0$ is a point in the interval.

Assume the Taylor series of $f$ at $0$ converges in the $\varepsilon$-neighborhood of $0$.
Denote by $\bar f$ its sum.
The monotonicity of $f^{(n)}$ gives a bound on the error $f(x)-\bar f(x)$ on one side from $0$; it follows that $\bar f(x)$ converges to $f(x)$ if $0<x<\varepsilon$ or $\varepsilon<x<0$, a contradiction.

It remains to consider the case when the Taylor series of $f$ at $0$ diverges in any neighborhood of $0$.
In this case, for any $\varepsilon >0$,
there is arbitrary large $n$ such that $|f^{(n)}(0)|>\tfrac{n!}{\varepsilon^n}$.
Applying monotonicity of $f^{(n)}$ and integrating, we get that $|f^{(k)}(x)|>2^n$ for any $k\le n$ and some $-4{\cdot}\varepsilon<x<4{\cdot}\varepsilon$, a contradiction.