A functional equation while attempting to integrate a function differently: $ f(x)-f(y) = z \bigl( f(xz) - f(yz) \bigr) $ Here is a functional equation :

For $f : \mathbb (0,\infty) \to \mathbb R$ continuous non-constant, $f$ has the property that for every $b,x,y>0$, $f(1) = -1$  and
$$
f(x) - f(y) = b \bigl( f(bx) - f(by) \bigr) \text .
$$

I want to show that the only possible solutions are $f(x) = a-\frac {a+1}x$ for some constant $a$, without using any form of calculus(with the exception of limits, if required).
For an approach, there are too many free variables, so I tried to set $b$ to a certain value, say $b=2$. Then I get:
$$
f(x)-f(y) =  2 \bigl( f(2x) - f(2y) \bigr) \text .
$$
but I don't know how to proceed : I can't use this to find a particular value of any single $x$, because the value of $x$ depends on the value of $2x$, and so on, so there's no way of proceeding I can think of. This also makes me feel that there is missing information, but that doesn't seem likely because in the presence of three constraining variables and domain, some kind of uniqueness must hold, right?
Setting $g(x) = xf(x)$ and trying to prove that it is constant has also proved futile, for the same reason as before.

Context : This post contains the question I am trying to answer.
Let $f : (0,\infty) \to \mathbb R$ be a function satisfying $f(x)-f(y) = \int_y^x \frac{\mathrm dt}{t^2}$ for all $y,x>0$. Under the fix $f(1) = -1$, we know that there is a unique function $f$ satisfying these conditions, given by $f(x) = -1 + \int_1^x \frac{1}{t^2}\,\mathrm dt = -\frac 1x$. I attempted to locate such an $f$ without using calculus, but rather using functional equations.
The logic was as follows : let $b>0$ be arbitrary, and in the expression $\int_x^y \frac{\mathrm dt}{t^2}$, make the change of variables $bt =u$, so that $b\,\mathrm dt = \mathrm du$. Accordingly, we get:
$$
f(x)-f(y) = \int_x^y \frac{\mathrm dt}{t^2} = \int_{bx}^{by} \frac{b^2 \,\mathrm du}{u^2} = b \int_{bx}^{by} \frac{\mathrm du}{u^2} = b \bigl( f(bx)-f(by) \bigr) \text .
$$
and thus the functional equation comes out. Now if I can use non-calculus techniques to prove that this functional equation is satisfied only by $-\frac 1x$ then I will have obtained the antiderivative of $\frac 1{x^2}$.
 A: In this case, more free variables is better: let's assume $$f(x)-f(y) = z(f(xz)-f(yz))$$ for $x,y,z>0$. Then,
$$f(x)-z\,f(xz)=f(y)-z\,f(yz),$$ meaning that expression depends only on $z$:
$$f(x)-z\,f(xz)=g(z).$$ Letting $x=1$, we see that $g(z)=-1-z\,f(z)$, i.e. we have
$$f(x)-z\,f(xz)=-1-z\,f(z).$$ Exchanging $x$ and $z$, we get
$$f(z)-x\,f(xz)=-1-x\,f(x),$$ multiplying the first equation by $x$ and the second one by $z$ and subtracting, we arrive at
$$x\,f(x)-z\,f(z)=-x+z+xz(f(x)-f(z)).$$ This equation has to hold for any $x,z>0$, substituting any constant value $z_0\neq1$ for $z$, we see that $x\,f(x)=ax+b$. Since $f(1)=-1$, this means $b=-a-1$, i.e. $x\,f(x)=ax-a-1$.
A: Suppose we have a functional equation
$$ x(f(x)-f(x\,y)) = z(f(z)-f(z\,y)) \tag{1}$$
which holds for all $\,x,y,z>0.\,$
We want solutions for $\,f(x)>0.$ Since this
equation holds for all $\,x,z>0\,$ both sides depend only on $\,y.\,$ call it $\,g(y).\,$
Thus if $\,z=1\,$ we define
$$ g(x) := f(1)-f(x). \tag{2} $$
Equations $(1)$ and $(2)$ imply that
$$ f(x) - f(x\,y) = g(y)/x. \tag{3} $$
Add equations $(2)$ and $(3)$ to get
$$ f(1) - f(x\,y) = g(x) + g(y)/x. \tag{4} $$
Rewrote the left side of equation $(4)$ using equation $(2)$ to get
$$ g(x\,y) = g(x) + g(y)/x. \tag{5} $$
Eliminate the denominator in equation $(5)$ by defining
$$ h(x) := x\,g(x). \tag{6} $$
Rewrite equation $(5)$ using equation $(6)$ to get
$$ h(x y) = h(x)y + h(y). \tag{7} $$
Define the limit
$$ c := \lim_{x\to 0} h(x). \tag{8} $$
The limit as $\,x\to 0\,$ of equation $(7)$ is
$$ c = c\,y + h(y). \tag{8} $$
Solve equation $(8)$ for $\,h\,$ to get
$$ h(y) = c - c\,y. \tag{9} $$
Using equations $(9)$, $(6)$, and $(2)$ to get
$$ f(x) = f(1) + c - c/x. \tag{10} $$
The original equation in the question was
$$ f(x)-f(y)=b(f(b\,x)-f(b\,y)). \tag{11} $$
Replacing $\,y\,$ by $\,x\,y\,$ and
$\,b\,$ with $\,z/x\,$ in this equation
is equivalent to our equation $(1)$.
EDIT: This version is simpler than my previous one
from Jan 22, 2021.
