Denote by $\mathbb{Z}_{>0}$ the set of positive integers. Plugging $x=1, y=a$ into (1) we get $f(1) \geqslant 1$. Next, by an easy induction on $n$ we get from (2) that
$$f(n x) \geqslant n f(x) \text { for all } n \in \mathbb{Z}_{>0} \text { and } x \in \mathbb{Q}_{>0}$$
In particular, we have
$$f(n) \geqslant n f(1) \geqslant n \quad \text { for all } n \in \mathbb{Z}_{>0}$$
From (1) again we have $f(m / n) f(n) \geqslant f(m)$, so $f(q)>0$ for all $q \in \mathbb{Q}_{>0}$. Now, (2) implies that $f$ is strictly increasing; this fact together with (4) yields
$$f(x) \geqslant f(\lfloor x\rfloor) \geqslant\lfloor x\rfloor>x-1 \quad \text { for all } x \geqslant 1$$
By an easy induction we get from (1) that $f(x)^{n} \geqslant f\left(x^{n}\right)$, so
$$f(x)^{n} \geqslant f\left(x^{n}\right)>x^{n}-1 \quad \Longrightarrow \quad f(x) \geqslant \sqrt[n]{x^{n}-1} \text { for all } x>1 \text { and } n \in \mathbb{Z}_{>0}$$
This yields
$$f(x) \geqslant x \text { for every } x>1 \text {. }$$
(Indeed, if $x>y>1$ then $x^{n}-y^{n}=(x-y)\left(x^{n-1}+x^{n-2} y+\cdots+y^{n}\right)>n(x-y)$, so for a large $n$ we have $x^{n}-1>y^{n}$ and thus $f(x)>y$.) Now, (1) and (5) give $a^{n}=f(a)^{n} \geqslant f\left(a^{n}\right) \geqslant a^{n}$, so $f\left(a^{n}\right)=a^{n}$. Now, for $x>1$ let us choose $n \in \mathbb{Z}_{>0}$ such that $a^{n}-x>1$. Then by (2) and (5) we get
$$a^{n}=f\left(a^{n}\right) \geqslant f(x)+f\left(a^{n}-x\right) \geqslant x+\left(a^{n}-x\right)=a^{n}$$
and therefore $f(x)=x$ for $x>1$. Finally, for every $x \in \mathbb{Q}_{>0}$ and every $n \in \mathbb{Z}_{>0}$, from (1) and (3) we get
$$n f(x)=f(n) f(x) \geqslant f(n x) \geqslant n f(x)$$
which gives $f(n x)=n f(x)$. Therefore $f(m / n)=f(m) / n=m / n$ for all $m, n \in \mathbb{Z}_{>0}$.

Comment. The condition $f(a)=a>1$ is essential. Indeed, for $b \geqslant 1$ the function $f(x)=b x^{2}$ satisfies (1) and (2) for all $x, y \in \mathbb{Q}_{>0}$, and it has a unique fixed point $1 / b \leqslant 1$.