Entropy notation: What does this mean?
The first thing you should realize to understand it is that the $H(Y)$ notation corresponds to the calculation of the information entropy for the states in $Y$, such that $H(Y) = -\sum\limits_{y \in Y} p(y) \log[p(y)]$.  With that in mind, we can interpret the expression and equate the terms.  You are correct that the three parts correspond to the probability of $0$, $e$, and $1$, where $e$ is the state of erased information.  Expanding it we get:

$H((1-\pi)(1-\alpha), \alpha, \pi(1-\alpha)) $
$\quad= -(1-\pi)(1-\alpha)\log[(1-\pi)(1-\alpha)] - \alpha\log(\alpha) - \pi(1-\alpha)\log[\pi(1-\alpha)]$

This gives us the full expression for calculating the entropy of the system composed of those three states.  Next, as we usually do when proving an equality, we cheat and look at the answer to see what form we need it in.  Your last term has all the $\pi$ terms clumped onto one side, and all the $\alpha$ terms clumped onto another (except for where they are also a coefficient).  Thus we will organize the log terms in this manner.  First we use the log rules to expand the products:

$ \quad= -(1-\pi)(1-\alpha)\log(1-\pi) - (1-\pi)(1-\alpha)\log(1-\alpha) $
$\quad\phantom{=} - \alpha\log(\alpha) - \pi(1-\alpha)\log(\pi) - \pi(1-\alpha)\log(1-\alpha)$

Then we sort all the log terms in the order of the final expression:

$ \quad = -\alpha\log(\alpha) - (1-\pi)(1-\alpha)\log(1-\alpha)  - \pi(1-\alpha)\log(1-\alpha) $
$ \quad\phantom{=} - \pi(1-\alpha)\log(\pi) -(1-\pi)(1-\alpha)\log(1-\pi)$

Then we simplify and factor:

$ \quad = -\alpha\log(\alpha) - (1-\alpha)\log(1-\alpha) + (1-\alpha) \cdot [ -\pi\log(\pi) - (1-\pi)\log(1-\pi) ]$

And this is equivalent to:

$$H(\alpha, 1-\alpha) + (1-\alpha) H(\pi, 1-\pi)$$

This can be interpreted as the left part being the entropy from erasing (probability $\alpha$) or not erasing, and the right part being, in the event it was not erased, the entropy from it being a 1 (probability $\pi$) or a 0.  It should also make some intuitive sense that this will be equal to the entropy from the distribution of final states of $0$, $e$ (erased), and $1$.