If you remove "perhaps at the cost of extending the initial [finite] set to a larger [finite] set of prime ideals" from your question, then the answer is yes for a fixed Krull domain $R$ if and only if $R$ is a UFD.  First, suppose that the answer is yes for $R$.  Then let $x$ be an element of $K$ such that $v_P(x) = 1$ for a fixed height one prime $P$ and $v_Q(x) = 0$ for all other height one primes $Q$.  It is clear, then, that $P$ is principal, generated by $x$.  (Check this.)  Since then all height one primes of $R$ are principal, $R$ is a UFD. Conversely, if $R$ is a UFD, then every height one prime  $P$ of $R$ is principal, say, generated by $x_P$, and then $x = x_{P_1}^{n_1} x_{P_2}^{n_2} \cdots x_{P_k}^{n_k}$ satisfies the required condition for $x$.

The clarified question is an interesting one. I doubt it has a positive answer, but I haven't made much progress on it.  Maybe it's useful to observe that the question can be rephrased in term of the group $D(R)$ of all divisorial, or $v$-closed, ideals of $R$ under $v$-multiplication, which is freely generated by the height one primes of $R$ (where $I^v = (I^{-1})^{-1}$ is divisorial closure and $v$-multiplcation is $(I,J) \longmapsto (IJ)^v$).  The question asks if, for all height one primes $P_1, P_2, \ldots, P_k$, there exists a nonzero ideal $K$ such that, for all fractional ideals $I$ in the subgroup $(P_1, P_2, \ldots, P_k)$ of $D(R)$ generated by the $P_i$, there exists a nonzero ideal $J$ that is $v$-coprime to $(P_1 P_2 \cdots P_k K)^v$ such that $(IJ)^v$ is principal.  In fact, one can restrict the fractional ideals $I$ to a single prime ideal $P_1$ and its inverse $P_1^{-1}$.  Equivalently, one can restrict the vectors $(n_i)$ to a single elementary unit vector and its negative.  Moreover, all of the "$v$"s can be eliminated if $R$ is a Dedekind domain.  Note that the class group of $R$ is the group $D(R)$  modulo the subgroup of all nonzero principal fractional of $R$.  Since we may assume that $R$ is not a UFD, it must have infinitely many height one prime ideals, and therefore $D(R)$ is a free group on infinitely many generators.

EDIT: I now think the OP's qualified statement is true.  Here is a purported proof.  For each $i $ from $1$ to $k$, there exist nonzero $x_i^+, x_i^- \in K$  such that $v_{P_i}(x_i^\pm) = \pm 1$,  $v_{P_j}(x_i^\pm) = 0$ for all $j \neq i$, and $v_{Q}(x_i^\pm) \geq 0$ for all $Q$ other than the $P_i$.   The height one prime $v$-factorizations of the $2k$ principal fractional ideals $(x_i^{\pm})$ all together involve only finitely many height one primes (including the $P_i$), so we may collect those into a finite set $\mathcal P$.  Then, for any $(n_1,  n_2, \ldots, n_k) \in \mathbb{Z}^k$, the nonzero element $x= (x_1^\pm)^{|n_1|} (x_2^\pm)^{|n_2|} \cdots (x_k^\pm)^{|n_k|}$ of $K$, with the sign  of each $\pm$ in $x_i^{\pm}$  matching the sign of $n_i$,  satisfies the desired condition of the OP.  (Note that the primes outside of $\mathcal{P}$ don't show up in the height one prime $v$-factorizations of $(x)$, and so $v_Q(x) = 0$ for all $Q$ not in $\mathcal P$.)