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 $a \notin P$ 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 finitely many exceptions to $v_P(x) = 0$ among your constraints in condition (1).
EDIT: @LaurentMoret-Bailly pointed out in the comments that the qualifying question with "perhaps..." probably meant that the primes added to the list ought not depend on $x$. In that more likely interpretation, I think it's not true, but I don't have an example. I'd first start by trying $\mathbb{Z}[\sqrt{-5}]$ and $\mathbb{Q}[X,Y,Z]/(Z^2-XY)$, and, if that doesn't work, then $\mathbb{Q}[X,Y,Z,W]/(ZW-XY)$. They have class group $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, and $\mathbb{Z}$, respectively. If those don't work, then it's likely true. Sorry, that's all I have for now.