I will use co-frames. If $\bar{\theta}_1$, $\bar{\theta}_2$ and $\bar{\theta}_3$ is the co-frame dual to $E_1$, $E_2$ and $E_3$, then the co-frame dual to $\lambda E_1$, $E_2$ and $E_3$ is:
$\theta_1 = \lambda^{-1} \bar{\theta}_1$, $\theta_2 = \bar{\theta}_2$ and $\theta_3 = \bar{\theta}_3$.
We then have:
$d \theta_1 + 2\lambda^{-1} \theta_2 \wedge \theta_3 = 0$,
$d \theta_2 +2 \lambda \theta_3 \wedge \theta_1 = 0$ and
$d \theta_3 + 2 \lambda \theta_1 \wedge \theta_2 = 0$.
From Cartan's first structure equation $d \theta_i + \sum_j \theta_{ij} \wedge \theta_j = 0$, we then deduce that
$\theta_{12} = -\lambda^{-1} \theta_3$, $\theta_{13} = \lambda^{-1} \theta_2$ and $\theta_{23} = (\lambda^{-1}-2\lambda) \theta_1$.
We then compute the curvature $2$-forms $\Omega_{ij} = d\theta_{ij} + \sum_k \theta_{ik} \wedge \theta_{kj}$. We obtain
$\Omega_{12} = \lambda^{-2} \theta_1 \wedge \theta_2$, $\Omega_{13} = \lambda^{-2} \theta_1 \wedge \theta_3$ and $\Omega_{23} = (4-3\lambda^{-2}) \theta_2 \wedge \theta_3$.
We then get that
$\frac{1}{2} TP_1(\theta) = \frac{1}{2\pi^2}(-\lambda^{-4}+2\lambda^{-2}-2)Vol_{SO(3)}$, where $Vol_{SO(3)}$ denotes the volume form of the round $SO(3)$ (i.e. corresponding to $\lambda = 1$).
Since the volume of the round $SO(3)$ is $\pi^2$, the formula for the Chern-Simons invariant of $g_\lambda$ now follows.
Edit 1: I will provide more details. Using the formula (I think $6.1$ in that Chern-Simons paper):
$$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right),$$
we get that
\begin{align*} \frac{1}{2} TP_1(\theta) = &\frac{1}{8 \pi^2} \left(-\lambda^{-1}\lambda^{-1}(\lambda^{-1}-2\lambda) \, \theta_3 \wedge \theta_2 \wedge \theta_1 \right. \\
&-\lambda^{-1} \lambda^{-2} \, \theta_3 \theta_1 \theta_2 + \lambda^{-1} \lambda^{-2} \, \theta_2 \wedge \theta_1 \wedge \theta_3 \\
&\left.\,+(\lambda^{-1} - 2\lambda)(4 - 3\lambda^{-2}) \, \theta_1 \wedge \theta_2 \wedge \theta_3\right) \\
& = \frac{1}{2 \pi^2}(-\lambda^{-3} + 2\lambda^{-1} - 2\lambda) \, \theta_1 \wedge \theta_2 \wedge \theta_3
\end{align*}
But remember the formulas
$\theta_1 = \lambda^{-1} \bar{\theta}_1$, $\theta_2 = \bar{\theta}_2$ and $\theta_3 = \bar{\theta}_3$.
We thus pick up an extra factor of $\lambda^{-1}$ when written in terms of the standard volume form on $SO(3)$. This is how I obtained:
$$\frac{1}{2} TP_1(\theta) = \frac{1}{2\pi^2}(-\lambda^{-4}+2\lambda^{-2}-2)\,Vol_{SO(3)}.$$
Finally, since the volume of $SO(3)$ is $\pi^2$, we get that
$$\Phi(g_{\lambda}) \equiv \frac{1}{2}(-\lambda^{-4} + 2 \lambda^{-2} -2) \equiv -\frac{1}{2} \lambda^{-4} + \lambda^{-2} \quad \text{(mod $\mathbb{Z}$)},$$
which is what Chern and Simons wrote, up to a minor algebraic manipulation. Yes, it is a tricky calculation!
Edit 2: I will explain how to obtain the connection $1$-forms $\theta_{ij}$. We should have
$$d\theta_1 + \theta_{12} \wedge \theta_2 + \theta_{13} \wedge \theta_3 = 0.$$
We have that $d\theta_1 + 2 \lambda^{-1} \theta_2 \wedge \theta_3 = 0$.
It can be shown that $\theta_{12} = h \theta_3$. Basically, if $\theta_{12}$ has a non-zero $\theta_1$ component, then this would lead to a non-zero $\theta_1 \wedge \theta_2$ component of $d\theta_1$, which is a contradiction. It can similarly be shown that $\theta_{12}$ has no $\theta_2$ component either.
A similar reasoning gives that:
$$\theta_{12} = h\, \theta_3 \text{ , } \theta_{31} = g \,\theta_2 \text{ and } \theta_{23} = f \,\theta_1.$$
We therefore have, using the first two equations in edit 2, that:
$$ -g - h = 2 \lambda^{-1} .$$
Using similar equations for $d\theta_2$ and $d\theta_3$, we get that:
$$ -f - h = 2 \lambda, $$
$$ -f - g = 2 \lambda. $$
Solving these $3$ linear equations in $f, g$ and $h$ yields
$$f = \lambda^{-1} - 2\lambda \text{ , } g = -\lambda^{-1} \text{ , } h = -\lambda^{-1}.$$
Therefore, we have that
$\theta_{12} = -\lambda^{-1} \theta_3$, $\theta_{13} = \lambda^{-1} \theta_2$ and $\theta_{23} = (\lambda^{-1}-2\lambda) \theta_1.$
Best Answer
The elements of the dual frame are not closed as one-forms. By definition, each of the functions $e_i^*(e_j)$ is constant. Hence looking at the definition of the exterior derivative, you see see that $de_i^*(e_j,e_k)=-e_i([e_j,e_k])$. This is just $-c^i_{jk}$, i.e. the negative of the usual structure constants of the Lie Algebra, which are characterized by $[X_j,X_k]=\sum_ic^i_{jk}X_i$. Otherwise put, the exterior derivative on left invariant forms restricts to a mam $\mathfrak g^*\to\bigwedge^2\mathfrak g^*$ which is the dual of the Lie bracket, viewed as a map $\bigwedge^2\mathfrak g\to\mathfrak g$.