Suppose that I have a (generic) function $g$ and arguments $a, b \in \mathbb{N}$. I know that $g$ satisfies the inequalities
$$1 < \frac{g(b)}{b} < \frac{g(a)}{a} < 2.$$
I also know that the following implications are true:
$$g(a) < g(b) \Longrightarrow \left\{a < b \Longleftrightarrow \frac{g(a)}{b} < \frac{g(b)}{a}\right\},$$
and
$$b < a \Longrightarrow \left\{g(b) < g(a) \Longleftrightarrow \frac{g(b)}{a} < \frac{g(a)}{b}\right\}.$$
Does it follow that
$$a < b \Longleftrightarrow g(a) < g(b) \Longleftrightarrow \frac{g(a)}{b} < \frac{g(b)}{a}?$$
(Note that I don't have information about whether $g$ is injective or not.)
We try to find a counterexample for your implication, by assuming $ a<b $ and $ g(a)>g(b)$. Both of this conditions render the two implications you gave as true, because both conditions on the left of your implications are false, hence the implications are true by definition.
Now our assumptions fit a proportional function like for example $g(x)=\frac{k}{x}$. With a little bit of trying to match your inequality chain from the beginning we get for example:
$$a=2;\quad b=3; \qquad g(x)=\frac{10}{2\sqrt{x}}$$
This fits all your requirements, but shows your implication that $a < b \Longleftrightarrow g(a) < g(b)$ is not true.
Hence, the answer to your question "Does it follow that ..." is "No".