Can $k[p,q]$ be dense in $k[x,y]$? where $p_xq_y-p_yq_x \in k^*$.

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Let $k$ be a field of characteristic zero. Let $p,q \in k[x,y]$ have invertible Jacobian, namely, $p_xq_y-p_yq_x \in k^*$ (partial derivatives). It can be shown that each of $k[x,y]$ and $k[p,q]$ is dense in $k[[x,y]]$, see this related question. Then in the induced topology, $k[p,q]$ is dense in $k[x,y]$. (More precisely, when considering $p,q \in k[[x,y]]$, we should further assume that each of $\{p,q\}$ have no constant term).

But what if I do not like to have infinite sums:

Is it possible to find for $k[x,y]$ a metric $d$ or a topology $\mathcal{T}$such that $k[p,q]$ is dense in $(k[x,y],d)$ or in $(k[x,y],\mathcal{T})$?

Remarks: (1) See also this question.

(2) It is clear the $k[p,q]$ is not dense in the metric $d(u,v):=2^{-l}$, where $u=\sum_{i=0}^{n}u_iy^i,v=\sum_{i=0}^{m}v_iy^i \in k[x,y]$, $u_i,v_i \in k[x]$, and $l$ is the minimal such that $u_l \neq v_l$.

(3) If I am not wrong, I can show that a positive answer to my above question implies that the two-dimensional Jacobian Conjecture is true; notice that the opposite implication is trivial: If the two-dimensional JC is true, then $k[p,q]=k[x,y]$, and then trivially $k[p,q]$ is dense in $k[x,y]$ in any topology.

Important edit: In the third remark above, the accurate statement is: "If I am not wrong, I can show that a positive answer to my above question with a metric such that $k[x,y]$ is complete, implies that the two-dimensional Jacobian Conjecture is true". If $k[x,y]$ is not complete in that metric or we only found a topology, then I do not know if this implies a positive answer to the two-dimensional Jacobian Conjecture.

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An induced topology on a subspace is still a topology in its own right -- you can talk about the topology on $k[x,y]$ induced from $k[\![x,y]\!]$ without mentioning power series. A basis for the topology is $\{f+(x,y)^n\mid f\in k[x,y],n\in\Bbb N_0\}$. The completion of $k[x,y]$ with respect to this topology is $k[\![x,y]\!]$, but you don't need to complete to talk about it. Then you can just take the subspace topology on $k[p,q]$.