Answer: $f = 1$ and $f = 1/x$.
It is easy to check that these are solutions, so we prove there are no other solutions.

*f = 1 if f is not injective:* Suppose $a > b > 0$ and $f(a) = f(b)$. Then we have
$$f(1 + a x) = f(a) f(x + f(a)) = f(b) f(x + f(b)) = f(1 + b x)$$
for any $x \in \mathbb{R}_{>0}$. Let $c = a/b > 1$ and $x = y/b$, then we get $f(1 + c y) = f(1 + y)$.
Taking $a = 1 + c y$, $b = 1 + y$, we get
$$f(1 + (1 + c y) z) = f(1 + (1 + y) z)$$
for all $z \in \mathbb{R}_{>0}$. Now, for $1 < t < c$, let $y = \frac{t-1}{c-t} > 0$ and $z = \frac{c-t}{c-1} > 0$.
Then $(1 + c y) z = t$ and $(1 + y) z = 1$, thus $f(1 + t) = f(2)$. Moreover, we have

$f(1 + c^n x) = f(1 + x)$ for any $n \ge 1$ by induction, thus increasing $n$, we see
that $f(x) = f(2)$ is constant for all $x > 2$.
Finally, for any $x \in \mathbb{R}_{>0}$, let $y = 2 + 1/x$, then $y + f(x) > 2$ and $1 + x y > 2$, thus $f(x) = f(1 + x y)/f(y + f(x)) = 1$.

$f = 1/x$ if $f$ is injective: Let $x > 1$. For $y = \frac{x-1}{x} > 0$, we have $1 + x y = x$, thus
$f(1 - 1/x + f(x)) = 1,$
since $f(x) > 0$. Let $d = f(2) - 1/2$. Then $f$ injective implies that $f(x) = d + 1/x$.
For $y = x$, we have $x > 1$, $y + f(x) > 1$, $1 + x y > 1$, hence
$$\left(d + \frac{1}{x}\right) \left(d + \frac{1}{x + d + \frac{1}{x}}\right) = d + \frac{1}{1 + x^2}.$$
Simplifying to a polynomial identity and considering the constant term, we see
that $d = 0$. It follows that $f(x) = 1/x$ for $x > 1$. Now let $y = f(x)$. Then
$f(2 f(x)) = f(1 + x f(x))/f(x)$ and thus
$$f\left(\frac{2}{x}\right) = \frac{x}{1 + x \frac{1}{x}} = \frac{x}{2}.$$
Therefore $f(x) = 1/x$ for $0 < x < 2$.