The continuous model of the problem suggests that the limit does depend on $f$ (and $u$), more precisely, it depends on how fast the parameter $f$ is suppressed in the expression whose limit you are taking;  $\lim y(nt, f_n)/n$ will depend on the limit of $nf_n$ as $n \to \infty$.  Only when this limit is zero does one get the proposed formula.

The associated differential equation is $Y' = 1/(1+Ae^{BY})$ where $A = e^z$ and $B=uf$. Its solution vanishing at 0 is $Y(t) = H^{-1}(t)$ where $H(t) = t + (A/B)(e^{Bt} - 1)$.  It does look like this matches the asymptotic behavior of your sequence for $y$ both in the large and small range.  For small $t$, the expansion $Y(t) = t/(1+A) + O(t^2)$ corresponds to your formula, but I think the answer is not quite that simple: you have to establish whether the expression whose limit is calculated belongs to the small regime where $Y(t)$ is approximately linear, or the large regime where $Y(t)$ is logarithmic, $Y(t)=O(\log(t))$.  The limit uses $n$ iterations so we want to know, as a function of $B \sim 1/n$, whether the transition between regimes happens at a point much larger than $1/B$.   However, it's easy to calculate that the ratio $H(t)/t$ moves away from 1 (the difference is larger than some constant independent of $B$) as soon as $Bt$ is of order 1 (i.e., bounded below by a given positive constant) and this would spoil the limit if the differential equation is a good model of the difference equation. 

To see the small-$y$ behavior directly in the difference equation, it can be expanded in powers of $y$. 

$y(t+1) - y(t) = 1/(1+A)  - y(AB)/(1+A)^2 + O(y^2)$

Your formula proposes that when $B \sim 1/n$, the effect of the $y^{\geq 1}$ terms is of order smaller than $n$ for $t \in [0,n]$.   The sum of the first $t$ terms in $y^1$ will be of order $t$, so only on a short interval, $t << n^{1/2}$.   The calculation with the differential equation suggests that $f_n = f/n$ is too large a parameter ; this calculation with the difference equation can be used to prove that $f_n = f/n^k$ is small enough for any $k > 2$.  Adding higher degree terms to the approximate difference equation would, I suppose, only get closer to the picture suggested by the differential equation. 

To prove rigorously the predictions from the differential equation you could try to control $y$ by trapping the sequence $y(n)$ between two trajectories of the ODE.  If simulations are consistent with a heuristically "wrong" formula it would be very interesting to sort out what the truth is.