[EDITED to exhibit $j$ as a rational function of $J_2,J_3,J_4$,
and to fix various local errors etc.]
The action of ${\rm SL_2} \times {\rm SL_2}$ on the $9$-dimensional space of
$(2,2)$ forms has a polynomial ring of invariants, with generators in degrees
$2,3,4$. If we write a general $(2,2)$ form $P(x_1,x_2;y_1,y_2)$ as
$(x_1^2, x_1 x_2, x_2^2) M_3 (y_1^2, y_1 y_2, y_2^2)^{\sf T}$ where
$M_3$ is the $3 \times 3$ matrix
$$
M_3 = \left( \begin{array}{ccc}
a_{00} & a_{01} & a_{02} \cr
a_{10} & a_{11} & a_{12} \cr
a_{20} & a_{21} & a_{22}
\end{array} \right) \; ,
$$
then $J_k$ ($k=2,3,4$) can be taken to be the $x^{4-k}$ coefficient of the
characteristic polynomial $\chi^{\phantom.}_{M_4}$ of the $4 \times 4$ matrix
$$
M_4 = \left( \begin{array}{cccc}
\frac12 a_{11} & -a_{10} & -a_{01} & 2 a_{00} \cr
a_{12} & -\frac12 a_{11} & -2 a_{02} & a_{01} \cr
a_{21} & -2 a_{20} & -\frac12 a_{11} & a_{10} \cr
2 a_{22} & -a_{21} & -a_{12} & \frac12 a_{11}
\end{array} \right) .
$$
This matrix is characterized by the identity
$$
P(x_1,x_2;y_1,y_2) =
(z_{11},z_{12},z_{21},z_{22}) M_4 (z_{22},-z_{21},-z_{12},z_{11})^{\sf T}
$$
where each $z_{ij} = x_i y_j$, together with the requirement that
$M_4$ has trace zero and becomes symmetric when its columns are listed
in reverse order and columns $2,3$ are multiplied by $-1$.
The invariants of degree $2$ and $3$ can also be written as
$$
J_2 = -\frac12 a_{11}^2 + 2(a_{01} a_{21} + a_{10} a_{12})
- 4 (a_{00} a_{22} + a_{20} a_{02}),
\quad
J_3 = -4 \det M_3;
$$
of course $J_4 = \det M_4$.
The (Jacobian of the) genus-$1$ curve $P=0$ is isomorphic with the
elliptic curve
$$
y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2.
$$
In particular this lets us compute the $j$-invariant of this curve
as a rational function of $J_2,J_3,J_4$:
$$
j = \frac{256 (J_2^2 + 12 J_4)^3}{16 J_2^4 J_4 - 4 J_2^3 J_3^2
- 128 J_2^2 J_4^2 + 144 J_2 J_3^2 J_4 + 256 J_4^3 - 27 J_3^4} \, .
$$
One way to obtain these results is as follows. First compute the
Hilbert series of the invariant ring. We find that it is
$1 / \bigl( (1-t^2) (1-t^3) (1-t^4) \bigr)$; this suggests
a polynomial ring of invariants with generators of degrees $2,3,4$,
and shows that if we find independent invariants $J_2,J_3,J_4$
of those degrees then ${\bf C}[J_2,J_3,J_4]$ is the full invariant ring.
Now use the basis $\{z_{ij}\}$ of the four-dimensional space,
call it $Z$, of sections of ${\cal O}(1,1)$; it is well-known that
$\{z_{ij}\}$ embeds ${\bf P}^1 \times {\bf P}^1$ into ${\bf P}^3$
as the quadric $z_{11} z_{22} = z_{12} z_{21}$,
identifying ${\rm SL_2} \times {\rm SL_2}$ with
the special orthogonal group ${\rm SO}(Q)$ where
$Q$ is the quadratic form $z_{11} z_{22} - z_{12} z_{21}$.
This identifies $P$ with some other quadratic form $\tilde P$
in the $z_{ij}$, determined uniquely modulo ${\bf C} Q$.
Now $Q$ is nondegenerate, so it identifies $Z$ with its dual $Z^*$,
and thus identifies quadratic forms on $Z$ with self-adjoint maps
$T: Z \to Z$, with $Q$ itself mapping to the identity map.
It is known that generically ${\rm SO}(q)$ orbits of such $T$
are determined by their spectrum, and thus by the characteristic polynomial
$\chi^{\phantom.}_T$. There is a unique translate $\tilde P + cQ$ of trace zero,
represented by the above matrix $M_4$. Hence the coefficients
$J_2,J_3,J_4$ of $\chi^{\phantom.}_{M_4}$ are invariant and independent,
as claimed.
To identify the elliptic curve, write $C$ as a double cover of
one of the ${\bf P}^1$'s by taking the discriminant of $P$ with respect to
the other ${\bf P}^1$, and then use classical formulas for the
Jacobian of a genus-$1$ curve $y^2 = {\rm quartic}$.
The formulas, though not pretty to look at, are short enough
to let us identify the coefficients with polynomials in $J_2,J_3,J_4$.
The resulting curve has a rational point whose $x$-coordinate is
a multiple of $J_2$; translating $x$ to put this point at $x=0$
yields the model $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$ exhibited above.
The visible rational point $(x,y) = (0,J_3)$
ought to correspond to the difference between the divisors
${\mathcal O(0,1)}_C$ and ${\mathcal O(1,0)}_C$,
but I haven't checked this.
