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 $x$.

If you don't 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 any Krull domain $R$.
Let $x = a/b \in K$ be nonzero, where $a,b \in R$ are nonzero.  Then $a$ does not lie in $P$ for all but finitely many height one primes $P$, and same with $b$, and thus $v_P(x) = 0$ for all but finitely many height one primes $P$.  Thus, if one has $v_{P_i}(x) = n_i$ for all $i$, then one automatically has $v_P(x) = 0$ for all but finitely many height one primes $P$, and therefore one can include the exceptions to that in your condition (1).