> But we always measure our weight using the unit kg! Are we right in doing so?

**No**. You are correct in thinking that we are **wrong** in doing so.

Actually, what is wrong with it, is our use of the word *weight*. If you instead of saying "My weight is 70 kg" said "My *mass* is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your *weight* $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your *mass* $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass (which should be constant nomatter where you are) that the scale shows on the screen will be 6 times too small as well:

$$W_{moon}=mg_{Earth} \implies \frac{1}{6}W_{Earth}=mg_{Earth} \implies \frac{1}{6}\frac{W_{Earth}}{g_{Earth}}=m$$

> What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a *gravitaional acceleration* with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$