Your Question is completely answered in the following book (Theorem 1.2, page 3). Also you can find some further related topics.
Pl. Kannappan, Functional Equations and Inequalities with Applications, Springer, 2009.
Theorem 1.2. Suppose $A : R\rightarrow R$ satisfies $A(x+y)=A(x)+A(y)$ with $c = A(1) > 0$. Then the following conditions are equivalent:
(i) $A$ is continuous at a point $x_{0}$.
(ii) $A$ is monotonically increasing.
(iii) $A$ is nonnegative for nonnegative $x$.
(iv) $A$ is bounded above on a finite interval.
(v) $A$ is bounded below on a finite interval.
(vi) $A$ is bounded above (below) on a bounded set of positive Lebesgue measure.
(vii) $A$ is bounded on a bounded set of positive measure (Lebesgue).
(viii) $A$ is bounded on a finite interval.
(ix) $A(x) = cx$.
(x) $A$ is locally Lebesgue integrable.
(xi) $A$ is differentiable.
(xii) A is Lebesgue measurable.
Like you wrote, we can show that $f$ is not injective and there is some $a$ such that $f(a) = 2$. Suppose towards a contradiction that there exists a $b$ with $f( b) < 2$.
$P(a, x)$ gives $f(x) = f(2x + a)$ so we can find a $c > a$, with $f(c ) = f(b)$ Let $y \in \mathbb{R}^+$ such that $$c + f(c)y = 2y + a$$
Then $P(c, y)$ gives $f(c) = 2$ which is a contradiction.
Similarly, suppose that there exists a $b$ with $f(b ) > 2$. Then we can find a $c < a$ with $f(c) = f(b)$ and a $y$ with $$c + f(c)y = 2y + a$$
and $P(c,y)$ gives the same contradiction.
Best Answer
$f(f(y))-f(x+y)\leq -xf(y)$.
As $xf(y)>0$, $f(f(y))-f(x+y)<0$.
Now, if $f(y_0)>y_0$ for some $y_0>0$, then we can put $x=f(y_0)-y_0$ and $y=y_0$ which yields $f(f(y_0))-f(f(y_0))<0$ which is clearly wrong.
Therefore, $$f(y)\leq y$$
Now, as $f(f(y))>0$, $xf(y)<f(x+y)\leq x+y$
$x(f(y)-1)<y$ $\forall$ $x,y>0$.
If $f(y)>1$ for any finite $y>0$, by making $x$ arbitrarily large, we obtain a contradiction.
Thus, $$f(y)\leq 1$$
As $f(f(y))>0$, $f(x+y)\leq 1$ and $xf(y)+f(f(y))\leq f(x+y)$, $$xf(y)<1$$
As $f(y)>0$, by making $x$ arbitrarily large, we obtain another contradiction.
Therefore, there exists no function $f:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that $xf(y)+f(f(y))\leq f(x+y)$.