Yes.

We can consider a sequence of disks $D_1,D_2,\ldots$ around $(\sqrt 2,\pi)$ such that the $n$th disk has precisely $n$ lattice points in its interior (and none on its boundary). Let $z_n$ be the lattice point in $D_n\setminus D_{n-1}$.

Construct accordingly a sequence $C_1, C_2, \ldots$ of simply connected compact sets with smooth boundary such that $C_{n+1}\setminus C_n$ is an annulus:
We let $C_1$ a a small closed disk around $f(z_1)$.
Given $C_n$, we can connect $f(z_{n+1})$ with $C_n$ via a straight line segment $\ell_n$ that does not pass through any other lattice point (and ends at its first intersection with $\partial C_n$. Then $C_n\cup \ell_n$ is simply connected, hence the complement is (via some map $\phi_n$) biholomorphic to $\{\,z\in\mathbb C:|z|>1\,\}$. Since almost all lattice points are "close" to $f(\infty)=\infty$, the distance $r_n$ between $f(\mathbb Z^2\setminus C_n)$ and $S^1$ is strictly positive.
Let $C_{n+1}=C_n\cup \phi_n^{-1}(\{\,z\in\mathbb C:|z|\le 1+\tfrac {r_n}2\,\})$.

Now to construct the desired homeomorphism $h$:

- Since $C_1$ and $D_1$ are disks, we can readily define $h\colon D_1\stackrel \approx \to C_1$
- Assume we have defined $h\colon D_n\stackrel\approx \to C_n$. To extend this, we need only find a homeomorphism between the closed annuli $D_{n+1}\setminus D^\circ_n$ and $\phi_n(C_{n+1}\setminus C^\circ_n)=\{\,z\in\mathbb C:1\le |z|\le 1+\frac{r_n}2\,\}$ that agrees with what we already have as homeomorphis between the inner boundaries. Viewing both sets as $[0,1]\times S^1$, we use the identity on the first factor and the given homeomorphism between the inner boundaries on the second factor.

The resulting map on $\mathbb R^2=\bigcup D_n$ is the desired homeomorphism if we can ensure that $\mathbb R^2=\bigcup C_n$.
Right now I am not sure if the construction above warrants this auttomatically. I suppose that one should be more "greedy", that is not only add $\ell n$ to $C_n$ (and then dilate it) but add something more while still maintaining simple connectivity at that step ...

Based on *Henning Makholm*'s and *Jim Belk*'s excellent comments, here's a complete rewrite of the above:

**Lemma.** Let $Q\subset \mathbb R^2$ be countable, closed, and discrete. Let $q_1,q_2,\ldots$ be an enumeration of $Q$.
Then there exists a homeomorphism $h\colon\mathbb R^2\to \mathbb R^2$ with $f(q_n)=(n,0)$ for all $n\in\mathbb N$.

*Proof.* Pick a point $a$ that is on none of the countably many lines through two points of $Q$.
Then for each $n$, the line segment $\ell_n$ from $a$ to $q_n$ is a compact set disjoint from $Q\setminus\{q_n\}$.

For $n\in\mathbb N$, $r>0$ let $C(n,r)$ be the convex hull of $\overline{B(a;r)}\cup\overline{ B(q_n;r)}$; this is the union of two disks and a rectangle, its boundary consists of two arcs and two line segments, it is star shaped around $a$, and every ray originating at $a$ intersects its boundary in exactly one point.

For $m\in\mathbb N$ let
$$ r(n,m) = \min\{\,|q_k-y|:k\ge m,k\ne n, y\in \ell_n\,\},$$
which exists and is positive because $\ell_n$ is compact and $Q$ (minus a finte set) is closed.
For $0<r<r(n,m)$ we have $C(n,r)\cap Q\subseteq\{q_1,\ldots,q_{m-1}\}\cup\{q_n\}$.
Clearly $m'>m$ implies $r(n,m')\ge r(n,m)$.
Also $r(n,m)\to\infty$ as $m\to\infty$ because $r(n,m)>r$ for all $m$ implies that $C(n,r)$contans infinitely many elements of $Q$, contradicting discreteness.

This allows us to pick for each $n$ a sequence $\{\rho_{n,m}\}_{m=n}^\infty$ that is strictly increasing and diverges $\to\infty$ as $m\to\infty$ and such that $\rho_{n,m}<r(n,m)$ holds for all $m\ge n$.
Now let
$$ C_n=\bigcup_{k=1}^n C(k,\rho_{k,n}).$$
Then $C_n$ is starshaped around $a$, compact, has a useer-friendly boundary consisting of finitely many arcs and line segments, and each ray originating in $a$ intersects the boundary in exactly one point.
Moreover, $C_n\subset C_{n+1}^\circ$ and $\bigcup_{n=1}^\infty C_n=\mathbb R^2$. We also have $C_n\cap Q=\{q_1,\ldots, q_n\}$ and $\partial C_n\cap Q=\emptyset$.
By linear interpolation between the boundaries we obtain a homeomorphism $h_0\colon \mathbb R^2\to\mathbb R^2$ that maps $a\mapsto 0$ and $\partial C_n\to nS^1$. Note that $h_0(q_n)$ is between $(n-1)S^1$ and $nS^1$. It is a simple task to deform each such annulus (and the central disk) in such a way that the boundary remains untouched and $f(q_n)$ moves to $(n-\tfrac12,0)$. With a final translation by $\frac12$ to the right, we obtain our desired homoeomorphism. $_\square$

Now the original problem is solved by applying the lemma to an enumeration $q_1,q_2,\ldots $ of $\mathbb Z\times Z$ and also to the enumeration $f(q_1),f(q_2),\ldots$.