Can this function be defined in a way to make it continuous at $x=0$?

There is an error at the end of your calculations. In the case $-1 < x < 0$ we have $$ \left\vert 1 - \frac{1}{x} \right\vert-\left\vert 1+\frac{1}{x}\right\vert= \left(1-\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)=2 $$ because the argument of the first absolute value is positive, and the argument of the second absolute value is negative.

A slightly simpler approach:

Since you are interested in the limit at $x=0$ it suffices to consider $x = (-1, 1)$, $x\ne 0$. For these arguments is $x-1 <0$ and $x+1> 0$, and therefore $$ f(x)=\frac{x}{\vert x-1 \vert - \vert x +1 \vert} = \frac{x}{-(x-1) - ( x +1)} = \frac{x}{-2x} =-\frac 12 $$ so that $$ \lim_{x \to 0 }f(x) = -\frac 12 \, . $$