I'm not quite sure how I should state this question.
This is one way:
What "natural" functions besides $g(x) = x$ and $g(x) =\dfrac{c^{x}}{c^{x}+c^{1/2}} $ satisfy $g(x)+g(1-x) = 1$?
By "natural" I mean that $g(x)$ is defined by a single expression for $0 \le x \le 1$, and not separately for $0 \le x \le \frac12$ and $\frac12 \le x \le 1$.
That is because, as I will show, if $g(x)+g(1-x) = 1$ then $\int_0^1 g(x) dx =\frac1{n+1}\sum_{k=0}^{n} g\left(\frac{k}{n}\right) =\frac12 $ for all $n$.
Another possible way is to ask if $g(x) =\dfrac{c^{x}}{c^{x}+c^{1/2}} $ then show that $\sum_{k=0}^n g(k/n) =\frac{n+1}{2} $.
This is inspired by my solution to Show that this sum is an integer.
First, I will show that if $g(x)+g(1-x) = 1 $, then $\int_0^1 g(x) dx =\frac1{n+1}\sum_{k=0}^{n} g\left(\frac{k}{n}\right) =\frac12 $ for all $n$.
If $G(n) =\sum_{k=0}^{n} g(k/n) $,
$\begin{array}\\ 2G(n) &=\sum_{k=0}^{n} g(k/n)+\sum_{k=0}^{n} g(k/n)\\ &=\sum_{k=0}^{n1} g(k/n)+\sum_{k=0}^{n} g((n-k)/n)\\ &=\sum_{k=0}^{n} (g(k/n)+ g((n-k)/n))\\ &=\sum_{k=0}^{n} 1\\ &=n+1\\ \text{so}\\ G(n) &= \frac{n+1}{2}\\ \end{array} $
If $I =\int_0^1 g(x) dx $,
$\begin{array}\\ 2I &=\int_0^1 g(x) dx+\int_0^1 g(x) dx\\ &=\int_0^1 g(x) dx+\int_0^1 g(1-x) dx\\ &=\int_0^1 (g(x)+g(1-x)) dx\\ &=\int_0^1 1 dx\\ &= 1\\ \text{so}\\ I &= \frac12\\ &= \frac{G(n)}{n+1}\\ \end{array} $
Now, I will show that if $g(x) =\dfrac{c^{x}}{c^{x}+c^{1/2}} $ then $g(x)+g(1-x) =1 $.
$\begin{array}\\ g(x)+g(1-x) &=\dfrac{c^{x}}{c^{x}+c^{1/2}}+\dfrac{c^{1-x}}{c^{1-x}+c^{1/2}}\\ &=\dfrac{c^{x}(c^{1-x}+c^{1/2})+c^{1-x}(c^{x}+c^{1/2})}{(c^{x}+c^{1/2})(c^{1-x}+c^{1/2})}\\ &=\dfrac{(c+c^{x}c^{1/2})+(c+c^{1-x}c^{1/2})}{c+c^{1/2}(c^{x}+c^{1-x})+c}\\ &=\dfrac{2c+c^{1/2}(c^{x}+c^{1-x})}{2c+c^{1/2}(c^{x}+c^{1-x})}\\ &= 1\\ \end{array} $
If $f(x)$ is any odd function whatsoever, then $g(x) = f(x-\frac12)+\frac12$ satisfies $f(x)+f(1-x)=1$. So for example, $g(x) = (x-\frac12)^3+\frac12=x^3-\frac{3 x^2}{2}+\frac{3 x}{4}+\frac{3}{8}$ and $g(x)=\arctan(x-\frac12)+\frac12$ and $g(x) = e^{x-1/2}-e^{1/2-x}+\frac12$ are all examples. Your second example is of this form as well, with $f(x) = \frac12(e^{x/2}-e^{-x/2})/(e^{x/2}+e^{-x/2})$.