= Solution
The given equation is a <Maurer-Cartan equation>, with the sign appropriate to the chosen invariant coframe. Write $\lambda=(\lambda_{ij})$ as an antisymmetric matrix of one-forms, so $d\lambda=\lambda\wedge\lambda$. Using the graded product rule for the <exterior derivative>,
$$
\begin{aligned}
d^2\lambda_{ij}
&=d\lambda_{ik}\wedge\lambda_{kj}-\lambda_{ik}\wedge d\lambda_{kj}\\
&=\lambda_{i\ell}\wedge\lambda_{\ell k}\wedge\lambda_{kj}
-\lambda_{ik}\wedge\lambda_{k\ell}\wedge\lambda_{\ell j}=0.
\end{aligned}
$$
The two sums coincide after exchanging their dummy indices $k,\ell$. This is also associativity of matrix-valued <wedge products of differential forms>.
To make the relation to the <Jacobi identity> explicit, let $\theta^a$ be any independent coframe and write $d\theta^a=-\frac12c^a{}_{bc}\theta^b\wedge\theta^c$. Its dual invariant vector fields satisfy $[E_b,E_c]=c^a{}_{bc}E_a$, by the <exterior derivative of a one-form evaluated on vector fields>. Evaluating $d^2\theta^a=0$ on three basis fields gives the cyclic coefficient identity
$$
c^e{}_{bc}c^a{}_{ed}+c^e{}_{cd}c^a{}_{eb}+c^e{}_{db}c^a{}_{ec}=0.
$$
These are exactly the components of $[[E_b,E_c],E_d]+[[E_c,E_d],E_b]+[[E_d,E_b],E_c]=0$. Thus \b[the structure constants obey the Jacobi identity].
For the real four-dimensional case, abbreviate $a=\lambda_{12}$, $b=\lambda_{23}$, $c=\lambda_{31}$, $d=\lambda_{43}$, $e=\lambda_{41}$, $f=\lambda_{42}$; here $d$ without an argument is a one-form, whereas $d(\cdot)$ denotes the <exterior derivative>. Directly expanding the given equations gives
$$
\begin{array}{lll}
d a=-b\wedge c-e\wedge f,&d b=-c\wedge a+d\wedge f,&d c=-a\wedge b-d\wedge e,\\
d d=-e\wedge c-b\wedge f,&d e=-f\wedge a+d\wedge c,&d f=-d\wedge b-a\wedge e.
\end{array}
$$
Define $\alpha_1^\pm=a\pm d$, $\alpha_2^\pm=b\pm e$, and $\alpha_3^\pm=c\pm f$. For example,
$$
d\alpha_1^+=-b\wedge c-e\wedge f-e\wedge c-b\wedge f
=-(b+e)\wedge(c+f).
$$
The minus combination and the other two cyclic combinations give
$$
\boxed{d\alpha_1^\pm=-\alpha_2^\pm\wedge\alpha_3^\pm,\quad
d\alpha_2^\pm=-\alpha_3^\pm\wedge\alpha_1^\pm,\quad
d\alpha_3^\pm=-\alpha_1^\pm\wedge\alpha_2^\pm.}
$$
The six forms are independent, since the original pairs are their half-sums and half-differences. Their dual vector fields $E_i^+,E_i^-$ consequently obey
$$
[E_i^\pm,E_j^\pm]=\epsilon_{ijk}E_k^\pm,\qquad [E_i^+,E_j^-]=0.
$$
Each triple is a three-dimensional <Lie algebra ideal> with the <SO(3) Lie algebra> brackets; the two ideals commute and span the original six-dimensional algebra. This proves the <real four-dimensional rotation algebra splitting>
$$
\boxed{\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3).}
$$
This is a Lie-algebra statement. Globally $SO(4)$ is a central quotient of $SU(2)\times SU(2)$; a direct product of the corresponding $SO(3)$ groups is not asserted.
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