A very elementary change of variable Let $f,g\in C^0(\mathbb R)$ and consider having an equation of the form
$$f(x)g\left(\frac{y-x}{\alpha}\right)=F(x,y),$$
where the right hand side is some function involving $x,y\in\mathbb R$. For the question at hand, the right hand side of the equality doesn't come into it at all; it is only to pose the above equality. What I am interested in, exclusively is the left hand side of the equality.

What is it, precisely and rigorously, which allows me to reconsider the left hand side, equivalently, as

$$f(x)g\left(\frac{y-x}{\alpha}\right)=f(y-\alpha x)g\left(x\right)?$$

On the one hand, it seems like we equate both arguments of the functions in the concatenation on the left hand side and rearrange, via,
$$x=\frac{y-x}\alpha\iff \alpha x=y-x \iff \alpha x - y=-x,$$
to obtain the desired change. But this feels like cheating. What is going on behind this change of variable which allows one to reconsider the left hand side in this way?
Your second displayed equation makes no sense since it uses the same letters $x$, $y$ for different things.

We have a coordinate transformation $\psi: \&gt;{\mathbb R}^2\to{\mathbb R}^2$ defined by
$$\psi:\&gt;(u,v)\mapsto\left\{\eqalign{x&amp;=v-\alpha u\cr y&amp;=v\cr}\right.\ ,\qquad{\rm resp.},\qquad\psi^{-1}:\&gt;(x,y)\mapsto\left\{\eqalign{u&amp;={y-x\over\alpha}\cr v&amp;=y\cr}\right.\quad.$$
We then want to express the function $F:\&gt;{\mathbb R}^2\to{\mathbb R}$ in terms of the new coordinates. This means that we consider the function $$G(u,v):=F\bigl(x(u,v),y(u,v)\bigr)=F\bigl(v-\alpha u, v\bigr)=f(v-\alpha u) g\left({v-(v-\alpha u)\over\alpha}\right)\ ,$$
so that we obtain 
$$G(u,v)=f(v-\alpha u) g(u)\ .$$
Nothing is going on behind the scenes: this is just different ways of writing the same thing.  In your initial expression $f(x)g((y-x)/\alpha)$ you have that $f$ and $g$ are both linear functions of $x$ (and, given you've equated it to $F(x,y)$ that g is also a linear function of $y$).  So it doesn't really matter how you express the argument of both functions so long as you don't change the nature of them -- whatever works best for your analysis of the problem can be picked.

It might be clearer if you look at it as a two-step process though.  The first step is to change the variable from $x$ to $z$:
$$ z = \frac{y-x}{\alpha} \Longleftrightarrow x=y-\alpha \cdot z $$
so that your expression (in $z$) becomes $f(y-\alpha z)g(z)$.  Then in the second step you rename $z$ as $x$ (this $x$ is now a 'new' $x$ which is used because you like $x$ as a variable, or you need $z$ for something else, or for whatever reason).