Using the Gell-Mann matrices for SU(3), can one construct a basis for SU(2)?
That is, is there a linear combination over the 8 Gell-Mann matrices:
$$ \sigma_x=\sum_{i=1}^8 a_i\lambda_i\\ \sigma_y=\sum_{i=1}^8 b_i\lambda_i\\ \sigma_z=\sum_{i=1}^8 c_i\lambda_i $$
such that $\sigma_x,\sigma_y$ and $\sigma_z$ are $3\times 3$ matrices with the following properties:
$$ \sigma_x\sigma_x=1\\ \sigma_y\sigma_y=1\\ \sigma_z\sigma_z=1\\ \sigma_x\sigma_y+\sigma_y\sigma_x=0\\ \sigma_x\sigma_z+\sigma_z\sigma_x=0\\ \sigma_y\sigma_z+\sigma_z\sigma_y=0\\ $$
Your question is a little ambiguous. If you are seeking three 3×3 hermitean traceless matrices which satisfy the same Lie Algebra as the Pauli matrices, than you know form elementary angular momentum theory that the eigenvalues of each 3d irrep matrix are 1,0,-1, and so you cannot satisfy $\sigma_x^2=1\!\!1$.
If, instead, you are just just looking for three 3×3 hermitean traceless matrices which satisfy your anticommutator conditions, this too is not achievable. To have $\sigma_x^2=1\!\!1$, you need, w.o.l.g., one eigenvalue 1 and a double eigenvalue -1 for $\sigma_x$, so, then, a characteristic polynomial $(n-1)(n+1)^2$, which has a non-vanishing quadratic term: impossible for a traceless 3×3 matrix.