### Solution:

#### Part 1: Domain and Axis of Symmetry

1. Finding the Domain:

Given the function $f\left(x\right)=\log _{a}(3-x)+\log _{a}(x+1)$, we need the arguments of the logarithms to be positive. Therefore, we have two inequalities:
- $3-x > 0 \Rightarrow x 0 \Rightarrow x > -1$

Combining these, we find the domain of $f(x)$ is $-1 0$ and $a \neq 1$, the value of $a$ that satisfies the given condition is $\boxed{a=2}$.