Existence of an $R$-basis with at least one unit in it? Let $F$ be a domain and let $R\le F$ be a subring such that $F$ is a free $R$-module of finite rank $n$.  
Question: Is there an $R$-basis $\lbrace e_1,...,e_n\rbrace$ of $F$ such that at least one of the basis elements is a unit in $F$ ? 
As an example consider $R = \mathbb{Z}$ and $F=\mathbb{Z}[\text{i}]$ where we can take the units $1,\text{i}$ as basis.  
 A: If $R$ is a dedekind domain or if each finitely generated projective  $R$-module is free, one can choose $e_1=1$. This is explained in the answers of Angelo and Florian of this question: 
Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$?.
However, the counter-examples given there don't apply since they aren't domains.
A: First some general remarks. You're asking whether $S/R$ is free as a $R$-module. If $S$ is a $R$-algebra which is free of finite rank, then the map $R \to S$ splits as a map of $R$-modules. This fact was already noticed by Florian Eisele in his answer to the other MO question.
Now for an explicit counter-example to your question. Consider the ring $R={\bf Z}[x,y,z]/(x^2+y^2+z^2-1)$. It is an integral domain. It is known that there exists a $R$-module $M$ which is not free such that $R \oplus M \cong R^3$. For a nice construction, see e.g. Keith Conrad's notes. Explicitly we can take $M=\{(f,g,h) \in R^3 : xf+yg+zh=0\}$. Note that we can embed $M$ in $R^2$ by $(f,g,h) \mapsto (f,g)$, and the cokernel $R^2/M$ is a torsion module, so there exists $F \in R \backslash \{0\}$ such that $F \cdot R^2 \subset M$.
Now, we would like to construct a $R$-algebra structure on $R \oplus M$. We can do this by considering the $R$-algbera $S_0 = R \otimes_{\mathbf{Z}} \mathcal{O}$ where $\mathcal{O}$ is an order of a cubic field $K$. It is an integral domain, since the polynomial $x^2+y^2+z^2-1$ is irreducible over any field of characteristic not $2$. Let $(1,\alpha,\beta)$ be a $\mathbf{Z}$-basis of $\mathcal{O}$. Embed $R \oplus M$ in $S_0$ by $(f,(g,h)) \mapsto f+g\alpha+h\beta$. This won't be a subring of $S_0$ in general, but $S=R \oplus FM$ is a subring of $S_0$ since $(FM) \cdot (FM) \subset F^2 S_0 \subset R \oplus FM$. So we have constructed an integral domain $S$ over $R$ such that $S/R \cong M$ is not free over $R$.
I don't know whether it's possible to find a counterexample where $R \to S$ splits as a map of rings, in other words where $S=R \oplus I$ where $I$ is an ideal of $S$.
