Yes, also if $f$ has to be irreducible. Elaborating on Dave Witte Morris' comment, I take $\alpha=3-2\sqrt{2}$, $\beta=2-\sqrt{3}$, both clearly algebraic units whose inverses are their Galois conjugates $\bar{\alpha}=3+2\sqrt{2}$, $\bar{\beta}=2+\sqrt{3}$. But now the inverse of any power $\alpha^n$ is $\bar{\alpha}^n=\overline{\alpha^n}$, again the Galois conjugate (if $n \neq 0$), and similarly for $\beta$. And a product $\alpha^m\beta^n$ ($m \neq 0$, $n \neq 0$) has three Galois conjugates $\bar{\alpha}^m\beta^n$, $\alpha^m\bar{\beta}^n$, $\bar{\alpha}^m\bar{\beta}^n$, one of which is its inverse and the other two are the inverses of each other.

So your set $S$ certainly contains $X=\{\alpha^m\beta^n| m \neq 0$ or $n \neq 0\}$. Now $\log \alpha$ and $\log \beta$ are $\mathbb{Q}$-linearly independent and so $\{m\log\alpha+n\log\beta|m,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$ (and removing $0$ doesn't change that). It follows that $X$ is dense in $\mathbb{R}_{>0}$. Since the inverse and the Galois conjugates of $-\gamma$ are the negatives of the inverse and the Galois conjugates of $\gamma$ respectively, the set $S$ also contains $-X$, which is dense in $\mathbb{R}_{<0}$. Your statement now follows.