Given $a\ge 0$ there is a unique monotonic function $f:\Bbb R\to\Bbb R$ satisfying $f(x+y)=f(x)f(y),\forall x,y\in\Bbb R, f(1)=a$ There are many answers on this site where it is proven that, if a function $f:\Bbb R\to \Bbb R$ satisfying the functional equation $f(x+y)=f(x)f(y),\forall x,y\in\Bbb R$ is continuous and nowhere $0,$ then $f(x)=a^x$ for some $a>0,$ but how can one prove a monotonic function $f$ can be uniquely determined by $f(x+y)=f(x)f(y)$ and $f(1)=a$ for a given $a>0$?
Some of the useful posts I've already seen regarding the problem with assumed continuity:

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*Overview of basic facts about Cauchy functional equation


*How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$ assuming $f$ is continuous and not zero


*If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere


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


*This answer
One thought that appeared to me was to use the following theorem at some point:

Let $I\subseteq\Bbb R$ be an open interval, $f:I\to\Bbb R$ a monotonic function on $I$ and $I'=f(I)$ an open interval. Then $f$ is continuous on $I$.

I thought, if I could prove if $f$ satisfying $f(x+y)=f(x)f(y),\forall x,y\in\Bbb R, f(1)=a>0$ and being monotonic implies $f(\Bbb R)$ is an open interval, that would imply $f$ is continuous and hence uniquely determined by the given conditions, however, it appeared more complicated than the initial problem.
How should I tackle this problem, given that $f$ is monotonic, without assuming $f$ is continuous?
 A: Assume WLOG that $f$ is non-decreasing (case $f$ is non-increasing is similar). Note that taking $y <0$ arbitrary and $x=1-y > 1$ we get $$ 0<a=f(1) = f(x+y) = f(x)f(y) $$
Since $f(x) \ge a > 0$ (due to monotonic behaviour) we have $f(y) > 0$, since $y<0$ was arbitrary and $f$ is monotonic, we have $f(t) > 0$ for any $t \in \mathbb R$.
Now, consider function $g=\ln \circ f$. We have equation $g(x+y) = g(x)+g(y)$. Moreover, since $g$ is monotonic (as composition of two monotonic functions), it is continuous at every point expect at most countably many (because sets $A_n = \{x \in \mathbb R: g(x) - \lim_{y \to x^-}g(y) > \frac{1}{n}\}$ is at most countable due to increasing behaviour of $g$).
In particular $g$ is continuous at one point $z \in \mathbb R$. Since $g(x+y) =g(x)+g(y)$, it is easy to note that $g$ must be continuous at every point.
By standard solution of Cauchy equation we get $g(x) = cx$ for some constant $c \in \mathbb R$. In particular $f(x) = \exp(cx)$ and since $f(1)=a$ we get $\exp(c)=a$ so that $f(x)=a^x$.
