$f(x) = x \forall x\in \mathbb{R}_{> 0}$ if $a = 1$, no such functions otherwise.

First, note that if $a < 0$, there exists no such function as for a fixed $y$, we can choose $x$ large enough such that the right hand side is negative. So, from now on assume $a > 0$. Let the original equality be $(*)$. First, observe that $f$ is injective; if $f(t) = f(u)$ for some $t, u > 0$, then plugging in $x = t$ and $x = u$ into $(*)$ gives us $at = au$, and since $a\neq 0$, this becomes $t = u$. Next, we change $y$ with $f(y)$. This becomes: $$ f(f(x) + f(y)) = ax + \frac{1}{f\left(\frac{1}{f(y)}\right)} \ldots(1) $$Interchanging $x$ and $y$ in $(1)$, we get that \begin{align*} ax + \frac{1}{f\left(\frac{1}{f(y)}\right)} &= ay + \frac{1}{f\left(\frac{1}{f(x)}\right)} \\ \frac{1}{f\left(\frac{1}{f(x)}\right)} - ax &= \frac{1}{f\left(\frac{1}{f(y)}\right)} - ay \end{align*}Hence, there exists a real number $c$ such that for all $x > 0$, $$\frac{1}{f\left(\frac{1}{f(x)}\right)} = ax + c\ldots (2)$$. This also means $c\geq 0$. Plugging this into $(1)$ gives us: $$ f(f(x) + f(y)) = ax + ay + c $$for all $x, y > 0$. Then, since $f$ is injective, we conclude that for any positive real numbers $x_1, x_2, y_1, y_2$ with $x_1 + y_1 = x_2 + y_2$, we have $f(x_1) + f(y_1) = f(x_2) + f(y_2)$. From this (left to exercise), we get that $f$ is linear. Let $m$ and $d$ be real numbers such that $f(x) = mx + d$ for all $x > 0$. Since $f(x) > 0$ for all $x > 0$, then $m\geq 0$, $d\geq 0$, and at least one of $m$ and $d$ are non-zero. Plugging into $(2)$, we get: $$ m\left( \frac{1}{mx+d}\right) + d = \frac{1}{ax+c} $$ Taking limit as $x$ goes to $\infty$, the right hand side approaches $0$, while the left hand side approaches $d$ (for both cases $m = 0$ and $m > 0$). Hence, $d = 0$, and it simplifies to $\frac{1}{x} = \frac{1}{ax+c}$. This now obviously gives us $c = 0$ and no solutions if $a \neq 1$. For the case $a = 1$, it can be checked back that the only possible solution is $f(x) = x$ for all $x\in \mathbb{R}_{> 0}$.