Let's consider the following three matrices: $$J_1=\begin{bmatrix} 0&0&0\\0&0&-1\\0&1&0\end{bmatrix}, J_2=\begin{bmatrix} 0&0&1\\0&0&0\\-1&0&0\end{bmatrix}, J_3=\begin{bmatrix} 0&-1&0\\1&0&0\\0&0&0\end{bmatrix}$$ The real value linear combination of this basis elements spans the Lie algebra of $SO(3)$. The complex value linear combination of this basis elements spans the Lie algebra of $SU(2)$.
Thus, it's true that the basis elements are the same (so of course they follow the same commutation relations) but it's also true that the vector space of the SU(2) Lie algebra is much bigger since is the complex linear combination. So why do we consider them the same Lie algebra?
The three matrices you exhibit do not span the Lie algebra of $\mathrm{SU}(2)$. The conditions defining a matrix $A$ in $\mathfrak{su}(2)$ are that $\bar{A}^t+A=0$ and $\mathrm{tr} A=0$, so a basis is given by $$ \left( \begin{matrix} i & \\ & -i \end{matrix} \right), \quad \left( \begin{matrix} & 1 \\ -1 & \end{matrix} \right), \quad \left( \begin{matrix} & i \\ i & \end{matrix} \right). $$ Notice that $\mathfrak{su}(2)$ is not a complex vector space, so you are not allowed to multiply by complex numbers. To see this, take the product $$ i \left( \begin{matrix} i & \\ & -i \end{matrix} \right) = \left( \begin{matrix} -1 & \\ & 1 \end{matrix} \right) =: B. $$ Then $B$ does not sit in $\mathfrak{su}(2)$, because $\bar{B}^t + B \neq 0$.