$$ E(aX + b) = aE(X) + b $$
$$ E(aX + b) = \sum_x(ax+b)p_X(x) = a\sum_xp_X(x) + \sum_xbp_X(x) = aE(X) + b$$
Could someone extrapolate on this proof a little, I'm a little confused why the constant b is included in a sum given it's not multiplied by the rv X. Or is that purely a formalism when dealing with an expectation.
Taking it one step at a time.
$$\begin{align} \mathsf E(aX + b) &= \sum_x(ax+b)p_X(x) &&{\text{definition of expectation}\\ \text{/ law of unconscious statistician}} \\[1ex] & = \sum_x \big(ax\,p_X(x) + b\,p_X(x)\big) && \text{distribution} \\[1ex] & = \sum_x ax\,p_X(x) ~+~ \sum_x b\,p_X(x) && \text{association} \\[1ex] & = a\sum_x x\,p_X(x)~+~ b\sum_x p_X(x) && \text{distribution} \\[1ex] & = a\sum_x x\,p_X(x)~+~ b && \text{law of total probability} \\[1ex] & = a\,\mathsf E(X) + b && \text{definition of expectation} \\[2ex]\therefore\quad\mathsf E(aX+b) ~&=~ a~\mathsf E(X)+b && \text{quad erat demonstrandum } \end{align}$$
$\blacksquare$
The step is just associating the series into two so that you can treat the terms separately. Then by distributing the constants out you can see that $b$ is multiplied by the total probability; demonstrating that $\mathsf E(b) = b$. That was a step your proof skimmed over; likely feeling that it was obvious.