If $D$ is a division ring and $V$ is a vector space over $D$, then a subring $R$ of $\mathrm{End}_D(V)$ is dense if for any $n$ linearly independent vectors $v_1,\ldots, v_n$ and any vectors $w_1,\ldots, w_n$, there exists $r\in R$ with $rv_i=w_i$ for $i=1,\ldots, n$. Jacobson's density theorem says that any primitive ring has a faithful representation as a ring of dense operators.
It is well known that the collection of all finite rank operators on $V$ is a dense ring (I don't require rings to be unital). Here is my question:
If a dense ring of operators on $V$ contains a finite rank operator, must it contain all finite rank operators?
This is obviously the case if $V$ is finite dimensional. I am primarily interested in the case that $V$ has countably infinite dimension. If it makes a difference, I would be willing to let $D$ be an algebraically closed field.