In this example $x_3$ and $x_4$ were the non-pivot variables and therefore used as the free variables. Is it possible to select any two variables as free variables and if not, is there a general rule/way to tell if the variables you want to use as the free variable can actually be used as free variables?
I attempted manipulate the solution so $x_1$ and $x_2$ are the free variables but it seems I cannot escape using at least one of $x_3$ or $x_4$ as the free variable as one equation has $x_3 - x_4$ and another has $x_4 - x_3$.
In addition to my comment, let us first look at your exercise. We can rewrite these 4 equations as $$-x_3+x_4=x_1-1$$ $$-x_3+x_4=1-x_2$$ $$x_3=x_3$$ $$x_4=x_4$$
Now let us assume that $x_1$ and $x_2$ are free. This means we can pick anything we want for them. Let's choose $x_1=4$ and $x_2=5$. This results in the system $-x_3+x_4=3$ with $-x_3+x_4=-4$ and this system has no solutions!. In order for your overall system to work, it turns out that $x_1$ and $x_2$ cannot be chosen arbitrarily, but for $x_3$ and $x_4$, it won't matter. Another example where you cannot arbitrarily choose your own choice of free variable is when one variable has been "assigned" a fixed value. example: $$x_1=4$$ $$x_2+x_3=5$$ $$0=0$$ You have a choice to pick $x_2$ or $x_3$ as your free variable, but $x_1$ is determined no matter what.