I am looking at a family of functions $f : [0, \infty) \rightarrow [-\infty, \infty)$ satisfying the following property: $$f(bx) - f(b(x-1)) > f(a(x+1)) - f(ax) \quad \text{for all $a, b \in \mathbb{R}^+$ and $x \in \mathbb{Z}^+$.}$$ Another way of phrasing the condition is: there are mutually disjoint intervals $I_1, I_2, I_3, \ldots$ and real numbers $y_1 > y_2 > \ldots$ such that for each $x \in \mathbb{Z}^+$, $$f(a(x+1)) - f(ax) \in I_x \subseteq [y_{x+1}, y_x] \quad \text{for all $a \in \mathbb{R}^+$}.$$ Note that since $I_1, I_2, I_3, \ldots$ are mutually disjoint, if $y_{x+1} \in I_x$ then $y_{x+1} \not\in I_{x+1}$ and vice versa.
One function satisfying this property is $\log$: for any fixed $x \in \mathbb{Z}^+$, $\log(a(x+1)) - \log(ax) = \log\left(\frac{a(x+1)}{ax}\right) = \log\left(\frac{x+1}{x}\right)$ is constant, and furthermore, $\log\left(\frac{x}{x-1}\right) > \log\left(\frac{x+1}{x}\right)$.
My question is this: Is there any other function satisfying this property other than functions of the form $f(x) = a \log x + b$?
My suspicion is that the answer is no, but I don't know how to prove it. I will explain my idea so far:
Note that the above condition requires $f$ to be strictly increasing. Hence, $y_0 > y_1 > \ldots > 0$ and so the intervals $I_i$ will eventually get very small, i.e. $y_i - y_{i-1} \rightarrow 0$ as $i \rightarrow \infty$.
On the other hand, let $g_x(a) = f(a(x+1)) - f(ax)$ for $a \in \mathbb{R}^+$ and $x \in \mathbb{Z}^+$. If there exists some $x \in \mathbb{Z}^+$, such that $g_x$ is not constant, then $I_x$ must contain more than one point, i.e. $\sup I_x - \inf I_x > 0$. Based on my observations on some functions, it seems that the intervals $I_1, I_2, I_3, \ldots$ do not get smaller fast enough, so that there will eventually be overlap between $I_x$ and $I_{x+1}$ for large enough $x$. (One function that I tested is $f(x) = \log x + \varepsilon \sin x$ where $\varepsilon$ is a small number like $\varepsilon = 0.01$.)
However, I haven't been able to prove this observation. I did find some relationship between $g_x$ and $g_{x+1}$ but it's not a strong enough relationship: $$g_{x+1}(a) - g_{x+1}\left(\frac{x-1}{x} \cdot a\right) = g_x\left(\frac{x+1}{x} \cdot a\right) - g_x(a).$$
So, is my suspicion true that the only functions satisfying the above property have the form $f(x) = a\log x + b$? If yes, how can I prove this? If not, what other function satisfies this property?