Simple manipulation with the given equation leads to
$$\begin{aligned} 1 + y f(x) - y f(x + y) - y^2 f(x) f(x + y) &= 1, \\ y f(x) - y f(x + y) &= y^2 f(x) f(x + y). \end{aligned}$$
After cancelling $y \neq 0$ another manipulation gives
$$\begin{aligned} f(x) - f(x + y) &= y f(x) f(x + y), \\ \frac{f(x) - f(x + y)}{f(x + y)} &= y f(x), \\ \frac{f(x)}{f(x + y)} - 1 &= y f(x), \\ \frac{f(x)}{f(x + y)} &= 1 + y f(x), \\ f(x + y) &= \frac{f(x)}{1 + y f(x)}, \\ \frac{1}{f(x + y)} &= (1 + y f(x)) \cdot \frac{1}{f(x)} = \frac{1}{f(x)} + y. \end{aligned}$$
(evidently, all the cancelled expressions are nonzero) and so
$$\frac{1}{f(x + y)} = y + \frac{1}{f(x)}$$
for any $x, y \in \mathbb{R}^+$. Hence
$$y + \frac{1}{f(x)} = x + \frac{1}{f(y)}.$$
Setting $y = 1$ we get
$$\frac{1}{f(x)} = x + \frac{1}{f(1)} - 1 = x + c, \quad \text{so} \quad f(x) = \frac{1}{x + c}$$
with a constant $c$. As $f(x) > 0$ for any $x > 0$, we have $x + c > 0$ for any $x > 0$ and therefore $c \ge 0$.
We can easily check that the function $f(x) = 1/(x + c)$ satisfies the given conditions for any $c \ge 0$:
$$\left(1 + \frac{y}{x + c}\right) \left(1 - \frac{y}{x + y + c}\right) = \frac{x + c + y}{x + c} \cdot \frac{x + y + c - y}{x + y + c} = \frac{x + c + y}{x + c} \cdot \frac{x + c}{x + y + c} = \frac{x + c + y}{x + y + c}.$$
But $\frac{x + c + y}{x + y + c} = 1$, so the condition is satisfied for all $x, y > 0$ and $c \ge 0$.

Put $x = 1$ and $f(1) = a > 0$. Then we have
$$(1 + a y)(1 - y f(y + 1)) = 1,$$
$$a y - y f(y + 1)(1 + a y) = 0,$$
$$f(y + 1) = \frac{a}{1 + a y}$$
Now put $y = 1$ and $f(x + 1) = a/(1 + a x)$ to get
$$(1 + f(x)) \left( 1 - \frac{a}{1 + a x} \right) = 1,$$
$$f(x) \cdot \frac{1 + a x - a}{1 + a x} = \frac{a}{1 + a x},$$
$$f(x) = \frac{a}{1 + a x - a} = \frac{1}{x + \frac{1}{a} - 1} = \frac{1}{x + c}.$$
As in the first solution, we have $c \ge 0$ and it is easy to check that such a function satisfies the required conditions.