$t=\exp(x-u) \quad u=\exp(x-t)$ for some $x =0..1$ Can $u \ne t$? From deeper discussion of this question I have the pair of equations 
$$ t = \exp(x - u) \\ 
   u = \exp(x - t) $$
Rearranging 
$$ t/u = \exp(t)/\exp(u) \qquad \text{ or }\\
   u/\exp(u) = t/\exp(t) $$
I'm looking at $0 \le x \le 1$ if this is simpler. My hypothese is that 
   $t=u$ is required.
Trying with arithmetic or geometric mean of $u$ and $t$ didn't lead easily to a solution, also the use of the taylor-series for $\exp()$ didn't give me the idea.     
I think however it must be somehow easy to be proved, but after a time fiddling with it the penny still didn't drop...          

Q: does the hypothese hold that for all $x$ in the interval $t=u$ is required?

 A: Rearranging, you get:
$$ \tag{1} te^u = e^x = ue^t $$
and from there
$$ \tag{2} \frac{t}{e^t} = \frac{u}{e^u} $$
The function $t\mapsto t/e^t$ increases from $t=0$ to $t=1$, but after that it starts falling again -- so you can find $t\ne u$ that satisfies $(2)$ quite fine, and from there use $(1)$ to compute $x=\log(t)+u$.
The only question is whether $x$ will end up in your range from $0$ to $1$.
Unfortunately it never does. I don't have a slick theoretical proof of this, but by numerical experiment, as $t$ increases from $1$ to $e$ (after which point there is clearly no hope because the $\log t$ term alone will be too large), the required $x$ seems to increase monotonically from $1$ to about $1.22$.
(For $t\ge 1$ solving for $u$ gives $u=-W(-t/e^t)$, so we can get Wolfram Alpha to plot $x$ as a function of $t$).
So there is no solution with $t\ne u$ given your constraints on $x$ -- but for every $x>1$ there is one.
A: I've got possibly an ansatz, but get stuck at the end.
$$ \begin{array} {rl|rl|l}
         t &= \exp(x-u)  &         u &= \exp(x-t) &  \implies \\
 t \exp(u) &= \exp(x)    & u \exp(t) &= \exp(x) & \\
\end{array}
$$ so we have
$$  t \exp(u) = u \exp(t) \qquad \gt 0  $$
See a contourplot for $\log(t)+u = \log(u)+t$ drawn by W/A:

Assume now, that $u=t+d$ then we get
$$ t \exp(t) \exp(d)  = (t+d)\exp(t) \\
 t  \exp(d)  = t+d \\
 t (1+d+d^2/2+ \cdots)  = t+d \qquad \text{ we know $t \gt 0$ so we can cancel}\\
  (1+d+d^2/2+ \cdots)  = 1+d/t \\
  d+d^2/2+ \cdots  = d/t \\
$$
Hmm. If $t \ge 1$ we have our contradiction and $d=0$ is required. But if $t \lt d$ I'm again stuck.... 
