Hardness of approximation: what decision problem is hard exactly?
Most likely what is meant is the following statement:

If $P \ne NP$, then there is no polynomial-time approximation algorithm for Set-Cover that achieves a $(1-\epsilon) \log n$ approximation factor.

Or, equivalently: the existence of a polynomial-time approximation algorithm for Set-Cover that achieves a $(1-\epsilon) \log n$ approximation factor implies that $P=NP$.



Your proposed formulation at the end of your question might look tempting, but if you want to understand how to formalize the claim, I probably wouldn't suggest formalizing it that way -- that formulation introduces unfamiliar technical issues of its own.  The problem you listed is a promise problem (it's not a decision problem).  Promise problems are weird and some of your intuition that you've built up won't work quite right for promise problems.   Promise problems are weird and counter-intuitive; if you want to build your intuition, often it's best to simply avoid them, rather than try to absorb the messy details involved in formalizing them.