I am stuck at proving the following identity $$ \sum_{k = 1}^{+\infty} \frac{ \prod_{i = 1}^4(b_i^k - b_i^{-k}) }{ (q^{k/2} - q^{-k /2})^2 } = \frac{ 1 }{ 2 }\sum_{\alpha_i = \pm 1}(\prod_{i = 1}^4 \alpha_i) E_2\Big[ \begin{matrix} 1 \\ \prod_{i = 1}^4 b_i^{\alpha_i} \end{matrix} \Big] \ , $$ where we define the twisted Eisenstein series for $\phi = e^{2\pi i \lambda}$ with $\lambda \in [0,1)$, $\theta \ne 1$, \begin{align} E_{k \ge 1}\left[\begin{matrix} \phi \\ \theta \end{matrix}\right] \equiv & \ - \frac{B_k(\lambda)}{k!} \\ & \ + \frac{1}{(k-1)!}\sum_{r \ge 0} \frac{(r + \lambda)^{k - 1}\theta^{-1} q^{r + \lambda}}{1 - \theta^{-1}q^{r + \lambda}} + \frac{(-1)^k}{(k-1)!}\sum_{r \ge 1} \frac{(r - \lambda)^{k - 1}\theta q^{r - \lambda}}{1 - \theta q^{r - \lambda}} \ . \end{align}
I was able to prove a similar equality, $$ \sum_{k = 1}^{+\infty} \frac{ \prod_{i = 1}^3(b_i^k - b_i^{-k}) }{ (q^{k/2} - q^{-k /2}) } = \frac{ 1 }{ 2 }\sum_{\alpha_i = \pm 1}(\prod_{i = 1}^3 \alpha_i) E_1\Big[ \begin{matrix} -1 \\ \prod_{i = 1}^3 b_i^{\alpha_i} \end{matrix} \Big] \ , $$ by using \begin{align} \sum_{n = 1}^{+\infty}\frac{x^n}{q^{- \frac{n}{2}} - q^{\frac{n}{2}}} = \sum_{n = 0}^{+\infty} \frac{x q^{n + \frac{1}{2}}}{1 - x q^{n + \frac{1}{2}}} \ , \end{align} and noticing that \begin{align} E_1\left[\begin{matrix} -1 \\ x \end{matrix}\right] = - B_1(1/2) + \sum_{r = 0}^{+\infty} \frac{x^{-1} q^{r + \frac{1}{2}}}{1 - x^{-1}q^{r + \frac{1}{2}}} - \sum_{r = 0}^{+\infty} \frac{x q^{r + \frac{1}{2}}}{1 - x q^{r + \frac{1}{2}}} \ , \end{align}
However, I have failed so far trying to apply a similar trick.