Sampling from specific random distribution on sets
I have a suggestion for an algorithm that might get close to what you want:

Pick $m$ from the distribution $\mu$ (which I will describe below).

Choose $S$ uniformly at random from all sets containing exactly $k-1$ integers in $[m+1,\dots,\min(m+w,n))$.

Add $m$ to $S$.  Output $S$.


So, how do we pick the distribution $\mu$ to make this algorithm pretty good?  I'll describe how, using linear programming.
Let $\mu_i$ denote the probability of $i$ according to the distribution $\mu$.  Then, we have
$$\begin{align*}
\Pr[x \in S]
&amp;= \sum_m \Pr[x \in S \mid m] \Pr[m]\\
&amp;= \mu_x + \sum_{m=x-w+1}^{x-1} {k-1 \over \min(w-1,n-m-1)} \cdot \mu_m.
\end{align*}$$
Notice that this is linear in the $\mu_i$'s.  Now, let's treat the $\mu_i$'s as variables to be solved for.  Introduce the linear inequalities
$$\left| \Pr[x \in S] - {k \over n}\right| \le t,$$
where $t$ is another variable.  Also add the equality $\mu_1 + \dots + \mu_{n-k}=1$ and $\mu_{n-k+1}=\cdots=\mu_{n-1}=0$ and the inequalities $\mu_i &gt; 0$ for all $i$.  We'll try to minimize $t$ subject to all of these inequalities.  These are linear inequalities in the variables $\mu_0,\dots,\mu_{n-w},t$, so you can find a solution in polynomial time using linear programming.  That gives you a distribution $\mu$ to use in the algorithm above.
This comes close to meeting your requirements.  It ensures the set $S$ output by the algorithm meets your conditions 1, 2, and 3.  It gets close to having $\Pr[x \in S] = k/n$; this is not exactly true, but it will be approximately true; and you can measure how large the error is when you solve the linear program, by inspecting the value of $t$.  Also, due to the inequality $\mu_i &gt; 0$, every possible set has a non-zero probability of occurring.  Of course, this is only a heuristic, and it probably is not the maximum-entropy distribution, so you'll have to decide whether it is good enough for your needs.
You mention that you are dealing with a very large value of $n$.  Given that, an additional heuristic would be to assume that $\mu_i$ is constant for all $i \in [3w,n-4w)$.  Intuitively, the endpoints are the problem cases, and there is a loose symmetry among the middle values that makes it feel reasonable that they'll end up having about the same probability.  So, we can enforce the constraint $\mu_{3w} = \mu_{3w+1} = \cdots = \mu_{n-4w-1}$, by replacing $\mu_i$ with $\mu_{3w}$ for all $i \in [3w,n-4w)$.  Now there are only about $6w$ variables, namely, $\mu_1,\dots,\mu_{3w},\mu_{n-4w},\dots,\mu_{n-w}$ and $t$.
Note that it is easy to express $\Pr[x \in S]$ as a linear sum of the $\mu_i$ variables; this sum can be computed in $O(w)$ time.  So, we end up with a linear program in about $6w$ variables, so you can apply a LP solver, and the resulting system will have $O(w)$ variables and inequalities instead of $O(n)$ variables and inequalities, so the LP solver should be a lot faster.