Suppose one has a set $S$ of positive real numbers, such that the usual numerical ordering on $S$ is a wellordering. Is it possible for $S$ to have any countable ordinal as its order type, or are the order types that can be formed in this way more restricted than that?

Yes, one can have any countable ordering. Indeed any countable totally ordered set can be embedded in $\mathbb{Q}$. Write your ordered set as $ \lbrace a_1,a_2,\ldots \rbrace $ and define the embedding recursively: once you have placed $a_1,\ldots,a_{n1}$ there will always be an interval to slot $a_n$ into. 


You can get any order type. Let's assume you can get all order types up to but not including alpha, using subsets of (0,1]. If alpha=beta + 1 then squash your representation of beta and add an extra point. If alpha is a limit ordinal, choose a sequence of ordinals that converges to alpha and put the first one into (0,1/2], the second into (1/2,3/4] etc. and the result will have order type alpha. 


To complete the picture (the obvious remaining part). If ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus wellordered subsets of R are exactly countable ordinals. 


Using wellorderings of positive reals is actually the standard way to construct an Aronszajn tree. 

