Isn't $\frac{ax+bi}{ax+bi}$ equal to $1$? Isn’t $\dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=\sqrt{-1}$, & $a$,$b$ & $x$ $\in$ $R$

Division in the field of complex numbers is defined in the following manner:
$\frac{ax+bi}{ax+bi}=\frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$\frac{ax+bi}{ax+bi}=\frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a=0$ or $x=0$ or both. Because $\frac{a^2x^2+b^2}{a^2x^2+b^2}$ is not defined when $b=0$ and either $a=0$ or $x=0$ or both.
