If $\mathcal{H}$ is a van der Corput set of positive integers, then the closure of $\{ g^h\mid h\in\mathcal{H} \}$ contains the identity element. This generalizes the statement in the original post, because $\{ P(n)\mid n>0 \}$ is a van der Corput set for $P\in\mathbb{Z}[x]$ without a constant term.

By Terry Tao's remark above, it suffices to show that for any $0<\epsilon\leq 1/2$ and for any $\theta_1,\dots,\theta_n\in\mathbb{R}$ there is $h\in\mathcal{H}$ such that $\|h\theta_1\|,\dots,\|h\theta_n\|<\epsilon$. We follow closely the proof of Theorem 9 in Chapter 2 of Montgomery: Ten lectures on the interface between analytic number theory and harmonic analysis. By the earlier results in the chapter, there is a cosine polynomial
$$ T(x) = a_0 + \sum_{h\in\mathcal{H}} a_h \cos 2\pi hx $$
with real coefficients, nonnegative values, $a_0<\epsilon^n$, and $T(0)=1$. Put
$$ f(x):=\max(0,1-\|x\|/\epsilon), $$
and consider the expression
$$ a_0 + \sum_{h\in\mathcal{H}} a_h f(h\theta_1)\dots f(h\theta_n). $$
It suffices to show that this expression exceeds $\epsilon^n$, because then $f(h\theta_1)\dots f(h\theta_n)\neq 0$ follows for some $h\in\mathcal{H}$. The Fourier expansion
$$ f(x) = \sum_{k\in\mathbb{Z}}\hat f(k) e(kx) $$
converges absolutely, hence the above expression equals
$$ \sum_{h\in\mathcal{H}\cup\{0\}} a_h
\left(\sum_{k_1\in\mathbb{Z}}\hat f(k_1) e(hk_1\theta_1)\right)\dots \left(\sum_{k_n\in\mathbb{Z}}\hat f(k_n) e(hk_n\theta_n)\right)=$$
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)
\sum_{h\in\mathcal{H}\cup\{0\}} a_h e(hk_1\theta_1+\dots+hk_n\theta_n)=$$
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)
\sum_{h\in\mathcal{H}\cup\{0\}} a_h \cos 2\pi h(k_1\theta_1+\dots+k_n\theta_n)=$$
$$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)T(k_1\theta_1+\dots+k_n\theta_n).$$
The term corresponding to $k_1=\dots=k_n=0$ contributes $\epsilon^n$, while all the other terms are nonnegative, hence we are done.