Can we assume that each clause in 3SAT contains at least one positive literal? That is, there is no clause with all negative literals (but it is fine that there is a clause with all positive literals).
You can assume this, in the sense that you can transform a 3SAT problem into this form with only polynomial size overhead. The idea is to to a Tseitin Transform on one variable in each clause with 3 negative literals.
So if you wanting to eliminate $\lnot x$, you introduce a fresh variable $y$ and add the clauses $ x \vee y$ and $\lnot y \vee \lnot x$. This is basically just encoding $\lnot x \iff y$. You can add a third dummy positive variable to ensure that these clauses also are not all negative.