I feel like I might have made a mistake on this question, and would appreciate some feedback from someone more experienced than me.
If $G$ acts on $S$, we write $S^G = \{s \in S: gs = s \, \forall \, g \in G\}$ for the invariant ring. Our 2 x 2 matrices $P_{ij}$ act on $\mathbb{C}[X,Y]$ by $X \longmapsto P_{11}X + P_{12}Y$ and $Y \longmapsto P_{21}X + P_{22}Y$, and extending these mappings to $X^i$ and $Y^i$. So, I want to calculate the invariant rings $(i): \mathbb{C}[X,Y]^{GL_2}$ and $(ii): \mathbb{C}[X,Y]^{SL_2}$. My arguments went like this:
$(i)$ Suppose we have some element of $\mathbb{C}[X,Y]^{GL_2}$, say $z = \sum \limits_{i,j \geq 0} A_{i j}X^i Y^j$. Take the matrix $\lambda I$, where $I$ is the identity matrix and $\lambda$ is large. Then $z$ must satisfy $\sum \limits_{i,j \geq 0} A_{i j}X^i Y^j = \sum \limits_{i,j \geq 0} A_{i j}\lambda^{i+j} X^i Y^j$; so clearly $A_{ij}$ must be zero unless $i=j=0$; thus our invariant ring is just $\mathbb{C}$.
$(ii)$ Now our matrices have to have determinant 1. Take the matrix with $A_{11} =A_{22} = 0$ and $A_{12} = \alpha$, $A_{21}=-1/\alpha$ (sorry I can't draw matrices in latex!) Then we must have $A_{ji} = (-1)^{j} \alpha^{i-j}A_{ij}$; so for $i \neq j$, clearly this can't be true for all $\alpha$: thus $A_{ij} = 0$ unless $i=j$. So any element of the invariant ring must be of the form $\sum_i A_i (XY)^i$. But then taking any upper triangular matrix with diagonal elements 1, don't we deduce again that the invariant ring is just $\mathbb{C}$?
The reason I am confused is that these were 2 consecutive problems in the same worksheet, and I didn't expect them to have the same invariant ring. If someone could point out where I've gone wrong or confirm my solutions, I'd be very grateful.
Your computations are both correct.
An alternative way to see this is the following: suppose that $f\in\mathbb C[X,Y]$ is invariant under $SL(2,\mathbb C)$. One can easily check that this implies that the function $$\phi:(x,y)\in\mathbb C^2\longmapsto f(x,y)\in\mathbb C$$ is invariant under $SL(2,\mathbb C)$, that is, for each $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb C)$ we have $$\phi(ax+by,cx+dy)=\phi(x,y), \qquad\forall x,y\in\mathbb C.$$ I claim that this implies that $\phi$ is constant: for example, if $(x,y)$, $(x',y')$ are two non-zero elements of $\mathbb C^2$, there exists a matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb C)$ such that $x'=ax+by$ and $y'=cx+dy$, so invariance means that $\phi(x',y')=\phi(x,y)$. This tells us that $\phi$ is constant on $\mathbb C^2\setminus\{(0,0)\}$ and, since $\phi$ is continuous, that $\phi$ is constant.
Now check that $\phi$ being constant implies that $f\in\mathbb C$.
Next, suppose that $f\in\mathbb C[X,Y]$ is invariant under $GL(2,\mathbb C)$. Since $SL(2,\mathbb C)\subset GL(2,\mathbb C)$, $f$ is also invariant under $SL(2,\mathbb C)$ and, by what we have just done, $f\in\mathbb C$.