This answer contains a list of all the $\Lambda$-ring structures that I have found. If anyone can find any others, they should edit this list.

• The two $\Lambda$-ring structures on $\mathbb A^1$ give three product $\Lambda$-ring structures on $\mathbb A^2 = \mathbb A^1 \times \mathbb A^1$.

• One of these product structures, $\psi_p(x)=x^p$ and $\psi_p(y)=y^p$, has a symmetry of order two switching $x$ and $y$. The ring of invariants of this symmetry is another structure.

• One can look at the ring of invariants of $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$ under the action of a subgroup of $GL_2(\mathbb Z)$, where $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$ sends $x$ to $x^ay^c$ and $y$ to $x^by^d$. If that subgroup is $D_3$, $D_4$, or $D_6$, one gets a $\Lambda$-ring structure on $\mathbb Z[x,y]$. These can also be viewed as the $K$-theory $\Lambda$-rings of the simple Lie groups $SL_3$, $SP_4$, and $G_2$, respectively.

That is a total of $7$ structures.

Here is a proof sketch that they are distinct: These can all be expressed as $\psi_p$-stable subrings of the $\Lambda$-ring $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$, with the induced $\Lambda$-ring structure. It is easy to check that they are not isomorphic as subrings. Thus it sufficies to recover the embedding into that ring.

Extend $\psi_p$ to $\psi_n$ the obvious way: so that $\psi_n \circ \psi_m = \psi_{nm}$. Then, for any element $\alpha\in \mathbb Z[x,x^{-1},y,y^{-1}]$, the sequence $\psi_n(\alpha)$ is a sum of geometric progressions and so satisfies a finite linear recurrence relation. So if $\alpha$ is a generic element of a $\Lambda$-subring $R$ of dimension $2$, the recurrence relation is defined over $R$, and the roots generate $\mathbb Z[x,y,x^{-1},y,y^{-1}]$. By picking a generic element, we can determine, purely from the $\Lambda$-ring structure on $R$, its embedding, and thereby distinguish the different rings.

This answer contains a list of all the $\Lambda$-ring structures that I have found. If anyone can find any others, they should edit this list.

• The two $\Lambda$-ring structures on $\mathbb A^1$ give three product $\Lambda$-ring structures on $\mathbb A^2 = \mathbb A^1 \times \mathbb A^1$.

• One of these product structures, $\psi_p(x)=x^p$ and $\psi_p(y)=y^p$, has a symmetry of order two switching $x$ and $y$. The ring of invariants of this symmetry is another structure.

• One can look at the ring of invariants of $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$ under the action of a subgroup of $GL_2(\mathbb Z)$, where $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$ sends $x$ to $x^ay^c$ and $y$ to $x^by^d$. If that subgroup is $D_3$, $D_4$, or $D_6$, one gets a $\Lambda$-ring structure on $\mathbb Z[x,y]$. These can also be viewed as the $K$-theory $\Lambda$-rings of the simple Lie groups $SL_3$, $SP_4$, and $G_2$, respectively.

That is a total of $7$ structures.

Here is a proof sketch that they are distinct: These can all be expressed as $\psi_p$-stable subrings of the $\Lambda$-ring $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$, with the induced $\Lambda$-ring structure. It is easy to check that they are not isomorphic as subrings. Thus it sufficies to recover the embedding into that ring.

Extend $\psi_p$ to $\psi_n$ the obvious way: so that $\psi_n \circ \psi_m = \psi_{nm}$. Then, for any element $\alpha\in \mathbb Z[x,x^{-1},y,y^{-1}]$, the sequence $\psi_n(\alpha)$ is a sum of geometric progressions and so satisfies a finite linear recurrence relation. So if $\alpha$ is a generic element of a $\Lambda$-subring $R$ of dimension $2$, the recurrence relation is defined over $R$, and the roots generate $\mathbb Z[x,y,x^{-1},y,y^{-1}]$. By picking a generic element, we can determine, purely from the $\Lambda$-ring structure on $R$, its embedding, and thereby distinguish the different rings.

EDIT: In fact, the first two constructions suffice to produce infinitely many $\Lambda$-rings! Consider the product ring $\mathbb Z[x,y]$ where $\psi_p(x)=x^p$ and $\psi_p(y)=2T_p(y/2)$. Then the subring $\mathbb Z[x,yx^n]$ is a sub-$\Lambda$ ring for each $n\geq 0$, and none of these are isomorphic. To check that they are non-isomorphic, one can recover $x$ as the unique solution to $\psi_2(x)=x^2$, then invert $x$, then recover $y$ as the unique solution to $\psi_2(y)=y^2-2$ that generates the ring along with $x$ and $x^{-1}$. Thus the embedding of the ring into $\mathbb Z[x,x^{-1},y]$ is unique, so the embedding into $\mathbb Z[x,y]$ is unique, and because they are distinct as subrings, they are distinct.

Similarly, inside $\mathbb Z[x,y]$ where $\psi_p(x)=x^p$ and $\psi_p(y)=y^p$, there is the subring $\mathbb Z[x^{n+1}y^n+x^ny^{n+1},xy]$ which is a sub-$\Lambda$-ring and depends on $n$ for similar reasons.

However these constructions are birationally equivalent to previous defined constructions. So perhaps it is better to consider this up to birational equvialence!