Let $M$ be a smooth (hausdorff, second countable) manifold of dimension $n\geq 2$ and consider a $C^\infty$ function $f:M\to\mathbb R$.
Q: Under what conditions is $S = M\setminus f^{-1}(0)$ is dense in $M$?
I think it can be safely said that a sufficient condition is given by $\text{d}f\neq 0$; a proof could employ the fact that $S$ is in that case a regular submanifold that can be parameterized in adapted coordinates $(x^1,\dots,x^n)$ by $x^n=0$. But do there perhaps exist weaker conditions under which the statement can be shown to hold as well? Of particular interest to me is the case where $\text{d}f\neq 0$ almost everywhere, or where the set $\text{d} f=0$ has empty interior.
Both of the weaker conditions you mentioned would suffice.
If $\{df=0\}$ has empty interior, then the set $\{df\neq 0\}$ is dense and open, and so you could apply your regular submanifold argument to conclude that $S$ is dense in $\{df\neq 0\}$, which in turn is dense in $M$, therefore by the transitivity of denseness, $S$ is dense in $M$.
$df\neq 0$ almost everywhere is strictly stronger than the preceding condition (given the continuity of $df$), so it would of course suffice as well.
To certain extent, this is the best you can hope for, in the sense that if $\{df=0\}$ does have interior, $f$ will be constant on some neighborhood, so there will be some value $t\in \mathbb R$ (not necessarily $0$) for which $M\backslash f^{-1}(t)$ is not dense.
So any condition on $f$ you come up with that is invariant under, say addition by a constant, will have to imply that $\{df=0\}$ is dense.