The polynomial you gave in the comments, $x_1y_2 - y_2 x_1$, is invariant under $SL_2$.

Proof: It's the determinant of

 $$ \begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}$$ and determinants are invariant under left multiplication by matrices of determinant $1$.

It indeed generates the ring of invariants. You can check this using representation theory (bidegree $s,k$ polynomials form the representation $\operatorname{Sym}^s \otimes \operatorname{Sym}^k$ of $SL_2$, and because $\operatorname{Sym}^j$ is irreducible this has one invariant if $s=k$ and $0$ otherwise) or by observing that any two nonzero matrices with the same determinant are equal up to the action of $SL_2$.

The same idea can be used to find the $SL_n$-invariants in the tensor product of $n$ copies of $k[x_1,\dots,x_n]$.

The polynomial you gave in the comments, $x_1y_2 - y_2 x_1$, after correcting the typo to $x_1 y_2 - x_2 y_1$, is invariant under $\operatorname{SL}_2$.

Proof: It's the determinant of  $$ \begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}$$ and determinants are invariant under left multiplication by matrices of determinant $1$.

It indeed generates the ring of invariants. You can check this using representation theory (bidegree $s, k$ polynomials form the representation $\operatorname{Sym}^s \otimes \operatorname{Sym}^k$ of $\operatorname{SL}_2$, and because $\operatorname{Sym}^j$ is irreducible this has one invariant if $s=k$ and $0$ otherwise) or by observing that any two nonzero matrices with the same determinant are equal up to the action of $\operatorname{SL}_2$.

The same idea can be used to find the $\operatorname{SL}_n$-invariants in the tensor product of $n$ copies of $k[x_1,\dotsc,x_n]$.