I'm not sure if there is a name for it, but the proof is modular arithmetic, combined with the following observation.
Lemma: Given a positive integer, $n$, the sum of the digits of $n$ has the same remainder upon division by $9$ that $n$ does.
Proof. Let us consider $3$ digit numbers for the sake of exposition. We could work in greater generality, but it would require a little bit more notation.
Because $10=9+1$, we will also have that $100=(9+1)(9+1)=9(9+1+1)+1$, and more generally, $10^n=9k_n+1$ for some integer $k_n$ depending on $n$. Thus, $10^n-1$ is divisible by 9 for every $n$.
Now, write a number whose digits are $n=[x][y][z]=100x+10y+z$. Then $n$ minute the sum of the digits of $n$ is $(100x+10y+z)-(x+y+z)=99x+9y$ is a multiple of $9$.Therefore, the remainder upon dividing $n$ by $9$ is the same as the remainder upon dividing $(x+y+z)$ by $9$.
So what the lemma tells us is that by adding the digits of a number (and adding those digits again if necessary), you will eventually get the remainder upon division by $9$. The last step we need is that, if we add two numbers, their remainders add, and if we multiply two numbers, their remainders multiply.
To see this, let us add $9a+b$ and $9c+d$ to get $9(a+c)+(b+d)$. We see that dividing this sum by $9$, we will get the same remainder as if we divided $(b+d)$ by $9$.
Similarly, if we multiply, $(9a+b)(9c+d)=9(9ac+ad+bc)+bd$, and if we divide this by $9$, we get the same remainder as if we had divided $bd$ by $9$.
To summarize, we have two things going on. We can add digits to find the remainder upon division by $9$, and we know that the remainder of a sum/product is the same as the sum/product of the remainers (assuming we divide by $9$ and take the remainder if we get something larger than $9$).
All this can be generalized. If you search for "modular arithmetic" you will find a lot of resources on this kind of math.
