How does Ohm disappear in noise/power spectral density unit?
The thermal noise power spectral density for a resistor is not defined as \$\frac{V^2}{R}\$, that is the power dissipated by a resistance \$R\$ with a voltage \$V\$ across it (and hence a current \$\frac{V}{R}\$ flowing through it). The thermal noise power spectral density for a resistor is \$4kTR\$, where \$k\$ is Boltzmann's constant and \$T\$ is the resistor temperature in Kelvin. The constant 4 at the start is obviously dimensionless. \$k\$ has units of \$J K^{-1}\$, \$T\$ is in Kelvin as we said before, and \$R\$ is in Ohms.
Combining these units in the equation yields units of \$J \Omega\$. However, the concept of power is simply described as a transfer of energy per unit time, or in other words \$P = \frac{E}{t}\$. So another way of writing Joules is Watt-seconds, \$W s\$. Recall that \$P = VI\$ and you'll get units of \$VAs\Omega\$, or volt-amp-second-ohms. If you then realise that Ohms are simply a ratio of volts to amps, you'll end up with volt-squared seconds, \$V^2 s\$. Finally, because the unit of Hertz expresses that something happens a certain number of times per second it follows that \$Hz = s^{-1}\$, and you end with the result that the resistor PSD is measured in \$\frac{V^2}{Hz}\$. It's common to take the square root of the original \$4kTR\$ to yield the voltage power spectral density \$\sqrt{4kTR}\$, which of course has units of \$\frac{V}{\sqrt{Hz}}\$.