About fractional iterations and improper integrals Let $g(x,0) = x$ and $g(x,t+1) = g(x,t) - \dfrac{1}{g(x,t)}$ for every real $t$.
From the fact
\begin{align}
\int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right)dx&=\int_{0}^{\infty}f\left(x-\frac{1}{x}\right)dx+\int_{-\infty}^{0}f\left(x-\frac{1}{x}\right)dx=\\
&=\int_{-\infty}^{\infty}f(2\sinh T)\,e^{T}dT + \int_{-\infty}^{\infty}f(2\sinh T)\,e^{-T}dT=\\ (collecting\space terms )
&=\int_{-\infty}^{\infty}f(2\sinh T)\,2\cosh T\,d T=\\
&=\int_{-\infty}^{\infty}f(x)\,dx.
\end{align}
as discussed here : Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?
I am tempted to conclude 
$\int_{-\infty}^{\infty}f(g(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$
for every real $t$.
SO I wondered about the general situation :
Let $H(x)$ be some real-analytic function such that :
$\int_{-\infty}^{\infty}f(H(x))=\int_{-\infty}^{\infty}f(x)\,dx.$
and the integral exists,
and define, 
$G(x,0) = x$ and $G(x,t+1) = H(G(x,t)) $ for every real $t$.
Does this Always imply that :
$\int_{-\infty}^{\infty}f(G(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$ ?
Related :
$\int_0^{\infty} A( f(B(x)) ) + C(x) ) dx = \int_0^{\infty} f(x) dx$
Show $\int_0^\infty f\left(x+\frac{1}{x}\right)\,\frac{\ln x}{x}\,dx=0$ if $f(x)$ is a bounded non-negative function
Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?

$$edit :$$
I avoided talking about uniqueness to avoid making things to complicated , but when uniqueness of the inverse Abel functions seems an issue then the truth of this conjecture might give an intresting uniqueness criterion for complex dynamics ?
Btw notice $x-\frac{1}{x}$ has its fixpoint at infinity.
I assume not all of the real fixpoints of $H(x)$ can be parabolic.
Also the inverse Abel function for $H$ will not use real parabolic fixpoints as the correct way of computation.
IN FACT probably $H(x)$ never has real fixpoints to avoid the PROBLEMATIC case of convergeance : $f(x) , f(H(x)) , f(H(H(x))) , ... $ $f$(fixpoint) = nonzero constant !!
Although that is not a proof.
I must give some credit to my mentor tommy1729 who talked about this last Friday. 


$$EDIT\space 2 :$$
Too adress sheldon's comment :
To compute $g(x,\frac{1}{2})$ I use the following method :
$g(x,\frac{1}{2}) = \lim_{n\space \to +\infty} g(\frac{g(x,-n)+g(x,-n+1)}{2},n)  $
More generally
$g(x,t) = \lim_{n\space \to +\infty} g((1-t)\space g(x,-n)+ t\space g(x,-n+1),n)  $


$$EDIT \space 3$$
It might be intresting to note that a neccessary condition for the conjecture to be true seems to be this below
$$\frac{d}{dt} \space \int_{-\infty}^{\infty}f(g(x,t))= 0$$
One can then use differentiation under the integral sign and the chain rule but it seems like an alternative statement rather then a step closer to a solution ?
Im still playing with this idea.

 A: I asked: What is $g(0,0.5)$?  The Op's equations suggests iterating the inverse of $x \mapsto x-1/x$.  Using the Op's notation, what is $g(0,-1)$?  There are two answers, 1 or -1, which lead to two different half iterates.  These two half iterates are not connected in the complex plane. Let's call the half iterate h(x). 
The formal solution I used started with the reciprocal of $z \mapsto z-1/z$, which is $z \mapsto \frac{z}{1-z^2}$.  This second mapping moves the fixed point from infinity to zero, and it has a parabolic fixed point with multiplier 1. Will Jagy posted an overview of the formal Fatou coordinate for a parabolic point at math overflow, I posted a pari-gp program that implements this on mathstack.
So, now we have $\alpha(f(z))=\alpha(z)+1$, where $f(z)=\frac{z}{1-z^2}$, and the relevant half iterate for the Op's question would be the reciprocal.  I can post the formal asymptotic series for $\alpha(z)$ if the Op is interested.
$$g(x,0.5) = h(x) = \frac{1}{\alpha^{-1}(\alpha(1/z)+0.5)}$$
Then $g(0,0.5)\approx \pm 1.12833$, depending on whether we start with 1 or with -1, for $g(0,-1)$. The two solutions are not connected.  There are two other solutions, which are complex valued; these four formal half iterate solutions are not connected in the complex plane. 
Lets focus only on the solution with $h(x)=g(0,0.5)\approx -1.12833$.  For large positive values of x, this solution leads to $h(x)\approx x$, which is exactly what we would like it to be.  $h(-1.12833)$ is near a simple pole, with $h(-1.1)\approx-89.55$.  
$$h(x-1/x) = h(x) - \frac{1}{h(x)}$$
We can use that equation to get the limiting value for arbitrarily large negative values of x, since $x-1/x$ gets arbitrarily large negative as x approaches zero, from the positive side.
$$\lim_{x \to -\infty} h(x) = h(0)-\frac{1}{h(0)} \approx -1.12833-\frac{1}{-1.12833}\approx-0.24206 $$
Unfortunately, the Op's "tempted" conclusion 
"$\int_{-\infty}^{\infty}f(g(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$" 
does not hold for the half iterate, since for large negative values of x, we have the half iterate goes to a small constant.
Here is an image of the two real valued half iterates of $x-1/x$, each of which has a simple pole, respectively near $\pm -1.12833$

Here is a blow up of the negative real axis, from -30 to -10.  As noted, the asymptotic limit as x gets arbitrarily large negative is -0.24206

For t=0.5, we have the half iterate.  Starting with x-1/x, there is a formal divergent Laurent series, that is asymptotic as x gets arbitrarily large.  This series defines four different analytic functions, depending on which quadrant of the complex plane we start iterating in.
$g(x,0.5)=h(x) \sim x -
\frac{1}{2x} +
\frac{1}{8x^3} +
\frac{-1}{16x^5}+
\frac{3}{128x^7}+
\frac{5}{256x^9}+
\frac{-59}{1024x^{11}}+
\frac{83}{2048x^{13}}+
\frac{3363}{32768x^{15}}+...$
A: In response to my first answer, Mick commented, "I was thinking about merging the 2 solutions but that does not work. I believe a variant of the idea might still work, not sure how yet."  This answer seeks to generate a function that is a function similar to Mick's function, $x-\frac{1}{x}$, and for which all of the fractional iterates behave the symmetrically, for large positive x, and for large negative x.  Tommy's comments at eretrandre inspired the following sequence of functions.
$f_2(x)=x-\frac{1}{2x}\;\;\; g_2(x)=f^{o2}(x)$
$f_n(x)=x-\frac{1}{nx}\;\;\; g_n(x)=f^{on}(x)$
$$\lim_{n \to \infty} f_n(x)=x-\frac{1}{nx}\;\;\; g(x)=f_n^{on}(x)$$
Initially, I used brute force, using a lot of computer cycles to estimate the limit, which is the Laurent series for the closed form $g(x,t)$ function below.  Then the the following function might be the function which Mick is looking for.  One can compare this function with the initial function Mick suggested $x-\frac{1}{x}$.
$g(x)= x -
\frac{1}{x}   -
\frac{1}{2x^{3}}   -
\frac{1}{2x^{5}}   -
\frac{5}{8x^{7}}   -
\frac{7}{8x^{9}}   -
\frac{21}{16x^{11}}  -
\frac{33}{16x^{13}}  -
\frac{429}{128x^{15}}  -
\frac{715}{128x^{17}}  -
\frac{2431}{256x^{19}}   ...$
This could be expressed as $f(x) = \frac{1}{g(1/x)}$, which leads to a formal solution with a surprisingly simple converging Abel function.   The Abel function for f(x) is: 
$$\alpha(z)=\frac{-1}{2z^2}\;\;\;\alpha(f(z))=\alpha(z)+1\;\;\;\alpha^{-1}(z)=-\sqrt{\frac{-1}{2z}}$$
And then working from the simple closed form Abel function, we get:
$f(z)=\alpha^{-1}(\alpha(z)+1) \;=\; \frac{z}{\sqrt{1-2z^2}}$
Finally, Mick's desired function in closed form would be as follows.  With a little algebra, we generate all of the fractional iterates of g(z) as well.  Then, using Mick's notation we have the desired g(z,t) function, which has all fractional iterates defined as:
$$g(z,t)=\frac{1}{f(1/z)} = \sqrt{z^2 - 2t} \; = \; z -
\frac{t}{z}   -
\frac{t^2}{2z^{3}}   -
\frac{t^3}{2z^{5}}   -
\frac{5t^4}{8z^{7}}   -
\frac{7t^5}{8z^{9}}   -
\frac{21t^6}{16z^{11}} - ...$$
As $|z|$ gets larger, this function rapidly converges towards Mick's original function.  Here is a graph of g(z,1) and g(z,0.5), from -5 to 5. Red is real and green is imaginary.  The smaller green half loop is the half iterate, and the larger green loop is the full iterate.  
In a comment, Mick writes: "How does this still relate to the OP since you admitted the integral no longer works?"  The interesting thing is that the integral does work for all finite values of n, at least for $\exp(-g_n^2)$.  So I was confused, and I thought a couple of graphs might help.  
Here is a really interesting plot of $\exp(-g(z)^2)$, along with the $\exp(-g_8(z)^2)$, from -5 to 5 at the real axis.  Here, $\int_{-\infty}^{+\infty}\exp(-g(z)^2) = e^2\sqrt{\pi}\;\;\;$ whereas $\int_{-\infty}^{+\infty}\exp(-g_8(z)^2) = \sqrt{\pi}\;\;$ -- which is perhaps surprising!  Notice that the integral of $\exp(-g_8(z)^2)$ is the same as the integral of Mick's $f=z-1/z\;\;\; \int_{-\infty}^{+\infty}\exp(-f^2(z)) =\sqrt{\pi}$
 
So what's going on here is that that every one of those poles in $g_8(x)$ or the pole in $f(x)$ has a really nasty singularity where all of the derivatives go to zero, and the function is no longer analytic, for $\exp(-f(z)^2)$.  Lets look at the plots of $\exp(-g(z)^2)$, along with the $\exp(-g_8(z)^2)$, again, from -5+0.5i to 5+0.5i. The integral $\int_{-\infty+0.5i}^{+\infty+0.5i}\exp(-g(z)^2=e^2\sqrt{\pi}$ since it is not path dependent. But the integral of $g_8$ is path dependent!  The integral of Mick's $\exp(-f(x)^2)$ function is also path dependent, due to the nasty singularity at z=0.  The two functions are converging towards each other in the complex plane as the iteration count gets larger, even though at the real axis you have all of these nasty singularities, and for any finite value of n, the integral remains $\sqrt{\pi}$

