# Non-strictly monotonic bijection $f:\Bbb R\to(0,+\infty)$ satisfying $f(a)=1, f(0)=1, f(x+y)=f(x)f(y)$

Is there any way of constructing a non-strictly monotonic bijective function $$f:\Bbb R\to(0,+\infty)$$ satisfying: $$f(x+y)=f(x)f(y), f(0)=1, f(1)=a>0\space$$

(without a Hammel basis for $$\Bbb R$$ over $$\Bbb Q$$)?

This question, without the condition that $$f$$ is not strictly monotonic, has already been asked many times, but I couldn't think of any discontinuous bijection from $$\Bbb R$$ to $$(0,+\infty)$$ with the properties above. I know that strict monotonicity implies $$f(x)=a^x,\space\forall x\in\Bbb R$$. One idea was to take some dense additive subgroup $$G\subset\Bbb R$$ and define $$f(x)=a^x,\space\forall x\in G$$, but then, as we require injectivity and $$f>0$$, the problem arises with $$f(\Bbb R\setminus G)$$. I found a related answer where it is proven that $$f$$ is either identically $$0$$ or $$f>0\space\forall x\in\Bbb R$$, but I couldn't use that answer to construct a function I'm looking for because we haven't learned about a Hammel basis in real analysis lectures yet. I also eliminated $$f(x)=\alpha x,\alpha\in\Bbb R$$ after realizing I couldn't fix one $$\alpha$$.

Is there any more elementary method I'm failing to see?