A function $f(x)$ is an real function and analytic in an open interval of $x$-axis or the whole $x$-axis. Is there only unique way to analytically extend it to the whole complex plane?
I know identity theorem for holomorphic functions, but it requires that two functions equal in an open and connected set, while here I only need they equal in an open interval of x-axis which is a closed set of complex plane.
Given any real-analytic function $f: (a, b) \to \mathbb{R}$ (allowing $(a, b)$ to be half-infinite or infinite) and a point $x_0 \in (a, b)$, one can compute the unique extension of $f|_{(x_0 - r, x_0 + r)}$ to the ball $B_r(x_0) \subset \mathbb{C}$ for any $r$ no larger than the radius of convergence of the Taylor series of $f$ at $x_0$ (and small enough that $(x_0 - r, x_0 + r) \subseteq (a, b)$) using the Taylor series of $f$. So, given any two complex-analytic extensions of $f$ to $\mathbb{C} \to \mathbb{C}$, they agree on such a ball and by the Identity Principle must agree everywhere. In this sense, the answer to your question is yes.
In general, however, a real-analytic function $\mathbb{R} \to \mathbb{R}$ need not admit an analytic extension to a map $\mathbb{C} \to \mathbb{C}$. For example, consider $$ f(x) = \frac{1}{1 + x^2} ; $$ it is certainly a real-analytic map $\mathbb{R} \to \mathbb{R}$, but it admits no complex-analytic extension to any domain containing $+i$ or $-i$.