Let $h:\mathbb R^{>0}\to \mathbb R^{\ge 0}$ be a smooth function, satisfying $h(1)=0$, and suppose that $h(x)$ is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$.
Let $s>0$ be a parameter, and define $ F(s)=\min_{xy=s,x,y>0} g(x,y), $ where $g(x,y):=h(x)+ h(y)$.
Question: Can the minimum be obtained at two essentially different points?
That is, suppose that $F(s)=g(x,y)=g(\tilde x,\tilde y)$, for some $x,y,\tilde x,\tilde y>0$ satisfying $xy=\tilde x \tilde y=s$. Is it true that
$$ (x,y)=(\tilde x,\tilde y) \, \, \, \text{ or } \,\, (x,y)=(\tilde y,\tilde x)?$$
By symmetry, we can assume W.L.O.G that $x \le \sqrt{s}$.
It is not hard to see that the minimum must be obtained at a point where $x, y \le 1$ (if $s \le 1$) or $x,y \ge 1$ (if $s \ge 1$). Thus, if $s \le 1$, then we have $x,y=\frac{s}{x} \le 1$, which implies $s \le x \le \sqrt{s}$.
Edit:
I tried to produce counter-examples by using $g$ which are invariant under some automorphism of the hyperbola $xy=s$. (Then the set of minimizers is closed under the operation of this automorphism). I couldn't find such an automorphism which preserve the special additive structure of $g$.
Here is a partial analysis of the question for local minima:
Set $\psi(x)=h(x)+h(\frac{s}{x})$. Then
$$\psi'(x)=h'(x)-h'(\frac{s}{x})\frac{s}{x^2}, \tag{1}$$
and
$$\psi''(x)=h''(x)+h''(\frac{s}{x})\frac{s^2}{x^4}+2h'(\frac{s}{x})\frac{s}{x^3}. \tag{2}$$
Now, suppose $x$ is a local minimum of $\psi$. Then, equations $(1),(2)$ imply that
$$ h'(x)=h'(\frac{s}{x})\frac{s}{x^2} \, \, , \, \, h''(x)+h''(\frac{s}{x})\frac{s^2}{x^4}+2\frac{h'(x)}{x} \ge 0\tag{3}. $$
Subquestion: Suppose that $x,y$ satisfy $(3)$. Does $x=y$ or $x=\frac{s}{y} $ hold?
Yes, it's possible. Define $$h(x)=\begin{cases} (x-1)^2 & x\in (0,2] \\ 2-(\frac{4}{x}-1)^2 & x\in [2+\epsilon,3] \end{cases} $$ for some small $\epsilon$. We'll deal with the values on other domains in a bit, but let's first see what this gets us. For $s=4$, we have $g(2,2)=2$ and $g(x,y) = h(x)+h(4/x) = 2-(\frac{4}{x}-1)^2+(\frac{4}{x}-1)^2 = 2$ for $x>2+\epsilon$. So in fact $g(x,y)$ is constant along the segment of the hyperbola $xy=s=4$ thus defined.
To finish the job, just smoothly interpolate $h$ on the interval $[2,2+\epsilon]$ in such a way that it is greater than $2-(\frac{4}{x}-1)^2$, and similarly finish it off with a smooth segment on $[3,\infty]$ which is increasing and greater than $2-(\frac{4}{x}-1)^2$. This ensures that the minimum is on the segment we defined in the first place, and here there are infinitely many distinct minima.
If you prefer isolated minima, you can add small (so as to keep the resulting function increasing), smooth bump functions to $h$ as defined above in all but a discrete set of points.