Let $a,b$ be two constant real numbers with $a\neq 0$. Can anyone give a special solution of the functional equation $f(ax+b)=f(x)+1$, where $f:\mathbb{R}\rightarrow \mathbb{R}$?
Note. It is a type of the Abel functional equations, and if $a=1$, then $f(x)=[\frac{x}{b}]$ is an its solution.
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In this solution, I drop the condition that $a\neq 0$. If $a\neq 1$, then there does not exist such $f$. Taking $x:=\dfrac{b}{1-a}$ in the functional equation yields $$f\left(\dfrac{b}{1-a}\right)=f\left(\dfrac{ab}{1-a}+b\right)=f\left(\dfrac{b}{1-a}\right)+1\,,$$ which is absurd.
If $a=1$, then we have $$f(x+b)=f(x)+1$$ for all $x\in\mathbb{R}$. If $b=0$, then no function works. If $b\neq 0$, then let $g(y):=f(by)-y$ for each $y\in\mathbb{R}$. Thus, $$\begin{align}g(y+1)&=f\big(b(y+1)\big)-(y+1)=f(by+b)-y-1\\&=\big(f(by)+1\big)-y-1=f(by)-y=g(y)\,.\end{align}$$ In other words, $$f(x)=g\left(\frac{x}{b}\right)+\frac{x}{b}\tag{*}$$ for some periodic function $g:\mathbb{R}\to\mathbb{R}$ with period $1$.
In conclusion, for real constants $a$ and $b$, there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(ax+b)=f(x)+1\text{ for every }x\in\mathbb{R}$$ if and only if $a=1$ and $b\neq 0$. When $a=1$ and $b\neq 0$, all solutions take the form (*). In particular, the solution $f(x)=\left\lfloor\dfrac{x}{b}\right\rfloor$ for all $x\in\mathbb{R}$ obtained by the OP arises from taking $$g(x):=-\left\{x\right\}\text{ for all }x\in\mathbb{R}\,.$$ Here, $\{t\}$ is the fractional part of $t\in\mathbb{R}$.
For fixed $a\in\mathbb{R}\setminus\{1\}$ and $b\in\mathbb{R}$, let now consider a function $f:\mathbb{R}\to\mathbb{R}$ which satisfies the functional equation $$f(ax+b)=f(x)+1\text{ for all }x\in\mathbb{R}\setminus\left\{\frac{b}{1-a}\right\}\,.\tag{#}$$ Firstly, we assume that $a=0$. Then, we see that $f(x)=f(b)-1$ for any $x\in\mathbb{R}\setminus\{b\}$. Thus, all functions $f:\mathbb{R}\to\mathbb{R}$ with the condition (#) are of the form $$f(x)=\begin{cases}c&\text{if }x=b\,,\\c-1&\text{if }x\neq b\,.\end{cases}$$
Secondly, we assume that $a>0$. Write $I^+:=\left(\dfrac{b}{1-a},+\infty\right)$ and $I^-:=\left(-\infty,\dfrac{b}{1-a}\right)$. For $x\in I^+$, we can see that $$x-\frac{b}{1-a}=a^t\text{ or }t=\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}$$ for some $t\in\mathbb{R}$. Thus, if $g_+(t):=f\left(a^t+\dfrac{b}{1-a}\right)$ for each $t\in\mathbb{R}$, then $$\begin{align}g_+(t+1)&=f\left(a^{t+1}+\frac{b}{1-a}\right)=f\Biggl(a\left(a^t+\frac{b}{1-a}\right)+b\Biggr)\\&=f\left(a^t+\frac{b}{1-a}\right)+1=g_+(t)+1\,.\end{align}$$ Therefore, if $h_+(t):=g_+(t)-t$, then $h_+:\mathbb{R}\to\mathbb{R}$ is periodic with period $1$. That is, $$f(x)=h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\text{ for all }x\in I^+\,.$$ We obtain a similar result for $x\in I^-$. Thus, $$f(x)=\begin{cases} h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ h_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ where $h_+,h_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $c\in\mathbb{R}$ is an arbitrary constant.
Finally, we are dealing with the case $a<0$. We rule out the case $a=-1$, since there does not exist a solution $f$ with the said property. This is because of the contradiction below when $a=-1$: $$f(x)=f\big(-(-x+b)+b\big)=f(-x+b)+1=\big(f(x)+1\big)+1=f(x)+2$$ for all $x\neq \dfrac{b}{2}$. From now on, we assume that $a\neq -1$.
Note that $$\begin{align}f\big(a^2x+(a+1)b\big)&=f\big(a(ax+b)+b\big)=f(ax+b)+1\\&=\big(f(x)+1\big)+1=f(x)+2\end{align}$$ for all $x\neq \dfrac{b}{1-a}$. Let $A:=a^2$, $B:=(a+1)b$, and $\phi(t):=\dfrac{1}{2}\,f(t)$ for all $t\in\mathbb{R}$. Then, $$\phi(Ax+B)=\phi(x)+1$$ for every $x\neq \dfrac{b}{1-a}=\dfrac{B}{1-A}$. Since $A>0$ and $A\neq 1$, we have by the previous section of this answer that $$\phi(x)=\begin{cases} \eta_+\left(\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}\right)+\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}&\text{if }x>\frac{B}{1-A}\,,\\ C&\text{if }x=\frac{B}{1-A}\,,\\ \eta_-\left(\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}\right)+\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}&\text{if }x<\frac{B}{1-A}\,, \end{cases}$$ where $\eta_+,\eta_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $C\in\mathbb{R}$ is an arbitrary constant. Therefore, $$f(x)=2\,\phi(x)=\begin{cases} 2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ 2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ where $c:=2C$.
Recall that $f(ax+b)=f(x)+1$ for $x\neq \dfrac{b}{1-a}$. For $x>\dfrac{b}{1-a}$, we have $$f(ax+b)=f(x)+1=2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{1}$$ Because $ax+b<\dfrac{b}{1-a}$ when $x>\dfrac{b}{1-a}$, we conclude that $$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{\ln|a|}\,.$$ Since $\dfrac{b}{1-a}-(ax+b)=|a|\,\left(x-\dfrac{b}{1-a}\right)$, we obtain $$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}+\frac{1}{2}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{2}$$ Equating (1) and (2), we conclude that $$\eta_+\left(t\right)=\eta_-\left(t+\frac{1}{2}\right)\text{ for each }t\in\mathbb{R}\,.$$ Let $h:\mathbb{R}\to\mathbb{R}$ be given by $$h(t)=2\,\eta_+\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$ Ergo, $h$ is periodic with period $2$ and $$h(t-1)=2\,\eta_-\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$ Hence, $$f(x)=\begin{cases} h\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ h\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}-1\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ for some real constant $c$ and for some periodic function $h:\mathbb{R}\to\mathbb{R}$ with period $2$.
It is not difficult to prove that the three results are indeed solutions to (#). I shall omit the proof of this part as an exercise.