In the following i will make use of the sub-multiplicative property of matrix norms, i.e. if $A,B$ are matrices such that $AB$ exists, then $||AB|| \le ||A|| \cdot ||B||$ where $||\cdot||$ is a matrix norm.
Suppose that $A$ is invertible. Let $X$ be any $n \times n$ matrix. Then $||AX-I||=||A(X-A^{-1})||=||A(XA-I)A^{-1}|| \le ||A(XA-I)|| \cdot ||A^{-1}|| \le ||A|| \cdot ||XA-I|| \cdot ||A^{-1}||$ and so $\frac{||AX-I||}{||XA-I||} \le ||A|| \cdot ||A^{-1}|| = \kappa(A)$, where $\kappa(A)$ is the condition number of $A$.
By a similar argument you can show that $\frac{||XA-I||}{||AX-I||} \le \kappa(A)$ as well.