Theorem: Let $n> 0 \in \mathbb Z.$ Let $p_n$ stand for the number of integer partitions of $n$ and let $k$ be the number of consecutive integers in a partition. Then $p_n + \sum_{k \ge 1}(-1)^k(p_{n - k(3k - 1)/2} + p_{n - k(3k + 1)/2}) = 0$.
Proof: $1 = \left(\sum_{k \ge 0}p_kx^k\right)\left(1 + \sum_{k \ge 1}(-1)^k\left(x^{k(3k - 1)/2} + x^{k(3k + 1)/2}\right)\right)$.
The power series on the right hand side of the equality in the proof above equals $1$ and so the coefficient of $x^n$ must equal $0$ except for that of $x^0$. That means $p_n = p_{n - k(3k - 1)/2} = p_{n - k(3k + 1)/2} = 0.$ That much of the proof I think I understand. When we expand the product in the proof we get a term $\sum_{k \ge 1}(-1)^k\left(p_nx^{k(3k - 1)/2} + p_nx^{k(3k + 1)/2}\right)$ in the sum (I think). With that said, how do we get the coefficients of $x^{k(3k + 1)/2}, \ x^{k(3k - 1)/2}$ to become $p_{n - k(3k + 1)/2}, \ p_{n - k(3k- 1)/2}$, respectively?
The coefficient of $x^n$ in the product
$$\left(\sum_{k\ge 0}p_kx^k\right)\left(\sum_{k\ge 1}(-1)^k\left(x^{k(3k-1)/2}+x^{k(3k+1)/2}\right)\right)$$
is a bit more complicated than you realized. It may be helpful to use different indices for the two summations:
$$\left(\sum_{\ell\ge 0}p_\ell x^\ell\right)\left(\sum_{k\ge 1}(-1)^k\left(x^{k(3k-1)/2}+x^{k(3k+1)/2}\right)\right)\,.$$
From this you can see that you get an $x^n$ term $(-1)^kp_\ell x^n$ when $\ell+\frac{k(3k-1)}2=n$ and another when $\ell+\frac{k(3k+1)}2=n$, i.e., when $\ell=n-\frac{k(3k\pm1)}2$. In other words, you get $x^n$ terms with coefficients $(-1)^kp_{n-k(3k-1)/2}$ and $(-1)^kp_{n-k(3k+1)/2}$ for each $k\ge 1$ (with the understanding that $p_m=0$ if $m<0$), and the coefficient of $x^n$ is therefore
$$\sum_{k\ge 1}(-1)^k(p_{n-k(3k-1)/2}+p_{n-k(3k+1)/2})\,.$$