The answer depends on the values of the constants $J^x$, $J^y$, and $J^z$. Here is what direct computation yields:
If $J^x=J^y=J^z=0$, so that $A=0$, then $B_1$ and $B_2$ span a $2$-dimensional abelian subalgebra.
If $J^x=J^y=0$, but $J^z\not=0$, then $A$, $B_1$, and $B_2$ span a $3$-dimensional abelian subalgebra (i.e., a maximal torus in ${\frak{su}}(4)$.
If $J^x = \pm J^y \not = 0$ and $J^z=0$, then they generate a $4$-dimensional subalgebra isomorphic to $\mathbb{R}\oplus {\frak{su}}(2)$.
If $J^x = \pm J^y \not = 0$ and $J^z\not=0$, then they generate a $5$-dimensional subalgebra isomorphic to $\mathbb{R}\oplus\mathbb{R}\oplus {\frak{su}}(2)$.
If $J^x \not= \pm J^y $ and $J^z=0$, then they generate a $6$-dimensional subalgebra isomorphic to ${\frak{su}}(2)\oplus {\frak{su}}(2)$.
If $J^x \not= \pm J^y $ and $J^z\not=0$, then they generate a $7$-dimensional subalgebra isomorphic to $\mathbb{R}\oplus {\frak{su}}(2)\oplus {\frak{su}}(2)$.
All of these follow by direct computation with matrices, which is made easier when you write them out by interchanging the second and fourth columns and rows, because then everything conjugated into the Lie subalgebra of $S\bigl(SU(2)\times SU(2)\bigr)$.