I am failing to see where the $\sigma$ is going in the below.
Given that normal distribution's pdf is:
$$p(x) = \frac{1}{\sigma \sqrt{2 \pi}}\exp \left( -\frac{(x-\mu)^2}{2 \sigma^2} \right)$$
and we set $\phi = \frac{x-\mu}{\sigma}$.
I obtain:
$$\Phi(\phi) = \frac{1}{\sigma \sqrt{2 \pi}}\exp\left( -\frac{1}{2}\phi^2 \right)$$
and yet, the correct result it appears, is:
$$\Phi(\phi) = \frac{1}{\sqrt{2 \pi}}\exp\left( -\frac{1}{2}\phi^2 \right).$$
Is there something there going on with the fact that $\sigma$ is $1$ for normal standard distribution, and they imply that fact in the equation?
And then a bonus question, how can you use just the $\phi$ to show that expected value and variance is 0 and 1, without using the distribution. I know that $\mathbb{E}[X]=\int_{\mathbb{R}}x f(x) dx$, but it is shown in my notes that mean of standard normal distribution is zero by simply using this: $\mathbb E\left[\frac{X-\mu}{\sigma}\right]$. I fail to understand this step.
I'm going to change your notation a bit. Let $X\sim N(\mu, \sigma^2)$, and let $$Y = \frac{X-\mu}{\sigma}.$$ Since this is one-to-one over the support, we can apply a one-to-one change of variable. Thus $X = \sigma Y+\mu$, and $$f_Y(y) = \frac{f_X(\sigma y+\mu)}{\left|\frac{dy}{dx}\right|} = \sigma\cdot\frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{1}{2}\left(\frac{\sigma y+\mu-\mu}{\sigma}\right)^2\right\} = \frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}y^2\right\}$$ which is what you are seeking. It is also possible to integrate $P(Y<y)$, but you can do that on your own.
Notice that this is the density of a a standard normal. Thus, we know that $$E[Y] = 0$$ and $$\text{Var}[Y] = 1.$$
In other words we have just shown that if $X\sim N(\mu, \sigma^2)$, then the transformation $$\frac{X-\mu}{\sigma}$$ yields a standard normal distribution $N(0,1)$. That is why $$E\left[\frac{X-\mu}{\sigma}\right] = 0,$$ immediately.