The answer is yes, simply because they are symmetric functions. However, it can even be done with nonnegative coefficients. The formula I present in Reduction formula for Schubert polynomials combined with the Littlewood-Richardson rule will work, though there's a more intuitive way (still requiring the Littlewood-Richardson rule).

Namely, we can represent the Schur polynomial as the sum of weights of semistandard tableaux of the appropriate shape (my reference for this is Fulton's book Young Tableaux). The sets of boxes with entries $3$ and $4$ will form a collection of skew tableaux, which we can rectify with jeu de taquin to obtain two tableaux of straight shape in each case: the remaining piece of the original tableau with entries $1$ and $2$, and the rectified tableau with entries 3 and 4. This gives us the result as a sum of products of monomials, and we can identify which pair of Schur polynomials each monomial comes from by the shape of the tableaux. We end up with a sum of the form
$$\sum_{\mu,\lambda}{a_{\lambda\mu}s_{\lambda}(x_1,x_2)s_{\mu}(x_3,x_4)}$$
To obtain the multiplicity $a_{\lambda\mu}$, fix a pair of semistandard tableaux of shape $\lambda,\mu$ obtained this way and count how many rectifications result in that (ordered) pair. After this we may specialize the variables, obtaining
$$\sum_{\mu,\lambda}{a_{\lambda\mu}s_{\lambda}(x_1,x_2)s_{\mu}(x_1,x_2)}$$
Since the polynomials are in the same sets of variables, the Littlewood-Richardson rule applies, so we end up with a sum of the form
$$\sum_{\mu,\lambda,\nu}{a_{\lambda\mu}c_{\lambda\mu}^{\nu}s_{\nu}(x_1,x_2)}$$
While this is a reasonable task in principle, the calculations become horrible pretty quickly. This is of course normal, as the Littlewood-Richardson rule itself is #P-complete.