First, when $x \neq z \equiv x \pmod{f(x)}$, we have $f(z) \equiv x \pmod{f(x)}$. Then substituting $x + f(x)$ and $f(x)$ for $x, y$ in the original equation gives
$$f(x + f(x) + f(x)f(x + f(x))) = x + f(x) + f(x + f(x))f(f(x))$$
Taking both sides mod $f(x)$, we obtain an important relation
$$f(x) \mid x f(f(x)) \quad (1)$$
In particular, we have
$$f(x + y f(x)) \mid (x + y f(x)) f(f(x + y f(x)))$$
that is,
$$x + f(x)f(y) \mid (x + y f(x))(x + f(f(y))f(x))$$
which is equivalent to
$$x + f(x)f(y) \mid f(x)^2(f(y) - y)(f(f(y)) - f(y)) \quad (2)$$
Suppose $f(1) \neq 1$. We claim that there exists a large number $a$ such that
$$(a, f(a)) = 1 \quad (3)$$
This is because, by a simple mathematical induction, we have
$$f(1 + f(1) + \cdots + f(1)^n) = 1 + f(1) + \cdots + f(1)^{n-1} + f(1)^{n+1}$$
and their greatest common divisor is
$$(1 + f(1) + \cdots + f(1)^n, f(1) - 1)$$
So as long as $n$ satisfies $(n + 1, f(1) - 1) = 1$, (3) will hold. Clearly $n$ can be taken very large,
and the corresponding $1 + f(1) + \cdots + f(1)^n$ is the $a$ we want.
Now, taking $x = a$ in equation (2), we have
$$a + f(a)f(y) \mid (f(y) - y)(f(f(y)) - f(y))$$
Hence for all $y$, we have
$$f(f(y)) = f(y) \quad (4)$$
Substituting $f(x)$ for $x$ in the original condition:
$$f((y+1)f(x)) = (f(y)+1)f(x) \quad (5)$$
Then substituting $f(y) - 1$ for $y$ (this is legal, since $1$ is not in the range of $f$) gives
$$f(f(y)f(x)) = (f(f(y) - 1) + 1)f(x)$$
Therefore there exists a constant $c$ satisfying
$$f(x) = c(f(f(x) - 1) + 1) \quad (6)$$
Taking $x = 1 + f(1)$ in the above equation gives
$$1 + f(1)^2 = c(f(f(1)^2) + 1)$$
Note that by using (4)
$$f(f(1)^2) = (f(f(1) - 1) + 1)f(1) = \frac{f(1)^2}{c}$$
Substituting back gives $c = 1$. Finally, suppose the smallest fixed point of $f$ is $k > 1$ (which exists by (3) and the well-ordering principle); then by (5) we get that $k - 1$ is also a fixed point, a contradiction.
Combining the above, we must have $f(1) = 1$, hence
$$f(x) = x, \forall x \in \mathbb{N}$$
Substituting back to verify, this clearly holds.