When can number rings be spanned (as $\mathbb{Z}$-modules) by units? Let $\mathcal{O}$ be the ring of integers in an algebraic number field.  Define $R \subset \mathcal{O}$ to be the set of all $\mathbb{Z}$-linear combinations of units.  Since the product of two units is a unit, the set $R$ is a ring.
Question : Under what circumstances do we have $R = \mathcal{O}$?
Of course, this holds for $\mathcal{O} = \mathbb{Z}$.  It is also clearly false for imaginary quadratic number rings aside from the Gaussian integers.  But it is true for all the real quadratic number rings I have played with.  Maybe it holds whenever $\mathcal{O}$ has infinitely many units?  This might be too optimistic...
In case the above has a wild or overly complicated answer, the following question might be easier.
Question : Under what circumstances is $R$ finite-index in $\mathcal{O}$ as an abelian group?
 A: As was already pointed out, the only imaginary quadratic fields with $R=\mathcal O$ are $\mathbb Q(i)$ and $\mathbb Q(\zeta_3)$. This follows pretty easily from the structure of $\mathcal O^\times$. On the other hand, for real quadratic fields $\mathbb Q(\sqrt{d})$, with $d \in \mathbb Z$ square-free, one has $R=\mathcal O$ in the following cases only:


*

*$d\not\equiv 1 \pmod{4}$ and either $d+1$ or $d-1$ is a perfect square.

*$d\equiv 1 \pmod{4}$ and either $d+4$ or $d-4$ is a perfect square.
This is was first proved by Belcher, Integers expressible as sums of distinct units, Bull. Lond. Math. Soc. 6 (1974), 66–68.
There are similar results for certain cubic and quartic fields but the problem for general number fields seems to be wide open. Nonetheless, an interesting positive result in this direction was obtained by Frei, On rings of integers generated by their units, Bull. London Math. Soc. 44 (2012), 167–182:

For every number field $K$ there exists a number field $L$ containing $K$ such that $\mathcal O_L = R_L$.

