No, there do not. $\textbf{First}$ solution:} Suppose the contrary. By setting $y=-1,0,1$ in succession, we see that the polynomials $1-x+x^2, 1, 1+x+x^2$ are linear combinations of $a(x)$ and $b(x)$. But these three polynomials are linearly independent, so cannot all be written as linear combinations of two other polynomials, contradiction. Alternate formulation: the given equation expresses a diagonal matrix with $1,1,1$ and zeroes on the diagonal, which has rank 3, as the sum of two matrices of rank 1. But the rank of a sum of matrices is at most the sum of the ranks of the individual matrices. $\textbf{Second}$ solution:} It is equivalent (by relabeling and rescaling) to show that $1 + xy + x^2y^2$ cannot be written as $a(x) d(y) - b(x) c(y)$. Write $a(x) = \sum a_i x^i$, $b(x) = \sum b_i x^i$, $c(y) = \sum c_j y^j$, $d(y) = \sum d_j y^j$. We now start comparing coefficients of $1 + xy + x^2 y^2$. By comparing coefficients of $1+xy + x^2y^2$ and $a(x)d(y) - b(x)c(y)$, we get $$\begin{align*} 1 &= a_id_i - b_i c_i \qquad (i=0,1,2)\\ 0 &= a_id_j - b_i c_j \qquad (i \neq j). \end{align*}$$ The first equation says that $a_i$ and $b_i$ cannot both vanish, and $c_i$ and $d_i$ cannot both vanish. The second equation says that $a_i/b_i = c_j/d_j$ when $i \neq j$, where both sides should be viewed in $\text{}$ (and neither is undetermined if $i,j \in \{0,1,2\}$). But then $$a_0/b_0 = c_1/d_1 = a_2/b_2 = c_0/d_0$$ contradicting the equation $a_0d_0 - b_0c_0 = 1$. $\textbf{Third}$ solution:} We work over the complex numbers, in which we have a primitive cube root $\omega$ of 1. We also use without further comment unique factorization for polynomials in two variables over a field. And we keep the relabeling of the second solution. Suppose the contrary. Since $1+xy+x^2y^2 = (1 - xy/\omega)(1 - xy/\omega^2)$, the rational function $a(\omega/y) d(y) - b(\omega/y) c(y)$ must vanish identically (that is, coefficient by coefficient). If one of the polynomials, say $a$, vanished identically, then one of $b$ or $c$ would also, and the desired inequality could not hold. So none of them vanish identically, and we can write $$\frac{c(y)}{d(y)} = \frac{a(\omega/y)}{b(\omega/y)}.$$ Likewise, $$\frac{c(y)}{d(y)}= \frac{a(\omega^2/y)}{b(\omega^2/y)}.$$ Put $f(x) = a(x)/b(x)$; then we have $f(\omega x) = f(x)$ identically. That is, $a(x) b(\omega x) = b(x) a(\omega x)$. Since $a$ and $b$ have no common factor (otherwise $1+xy+x^2y^2$ would have a factor divisible only by $x$, which it doesn't since it doesn't vanish identically for any particular $x$), $a(x)$ divides $a(\omega x)$. Since they have the same degree, they are equal up to scalars. It follows that one of $a(x), xa(x), x^2a(x)$ is a polynomial in $x^3$ alone, and likewise for $b$ (with the same power of $x$). If $xa(x)$ and $xb(x)$, or $x^2 a(x)$ and $x^2 b(x)$, are polynomials in $x^3$, then $a$ and $b$ are divisible by $x$, but we know $a$ and $b$ have no common factor. Hence $a(x)$ and $b(x)$ are polynomials in $x^3$. Likewise, $c(y)$ and $d(y)$ are polynomials in $y^3$. But then $1 + xy + x^2 y^2 = a(x)d(y) - b(x) c(y)$ is a polynomial in $x^3$ and $y^3$, contradiction. $\textbf{Note:}$ The third solution only works over fields of characteristic not equal to 3, whereas the other two work over arbitrary fields. (In the first solution, one must replace $-1$ by another value if working in characteristic 2.)