Which is the mathematical theory that talks about these structures? Let's define $\sigma(n)$ as the sum of the digits of the integer $n$ modulo $9$, having posed that $\sigma(9) = 9$.
Now consider 2 number $a$ and $b$ in the set $\{1, \cdots, 9\}$. Which is the value of $\sigma{(ab)}$?
Starting from this problem, one can imagine to build a ``pythagorean table'' for $\sigma(ab)$. That is:
$$ 
\begin{array}{ccccccccccc}
   &   & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
   &   &   &   &   &   &   &   &   &   &   \\
 1 &   & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
 2 &   & 2 & 4 & 6 & 8 & 1 & 3 & 5 & 7 & 9 \\
 3 &   & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
 4 &   & 4 & 8 & 3 & 7 & 2 & 6 & 1 & 5 & 9 \\
 5 &   & 5 & 1 & 6 & 2 & 7 & 3 & 8 & 4 & 9 \\
 6 &   & 6 & 3 & 9 & 6 & 3 & 9 & 6 & 3 & 9 \\
 7 &   & 7 & 5 & 3 & 1 & 8 & 6 & 4 & 2 & 9 \\
 8 &   & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 9 \\
 9 &   & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\
\end{array}
$$
This is the first structure I'm looking for the right mathematical theory. In particular, I think it's something like an algebra on the set $\{1, \cdots, 9\}$, with the following product:
$$ a \times b = \sigma(ab)$$
Now, what can we say about the ``new'' product when we use $2$-digits numbers? 
Let's $n = 10a + b$ and $m = 10c + d$, with $a$, $b$, $c$ and $d$ are number in the set $\{1, \cdots, 9\}$. Then:
$$nm = 100ac + 10(b+d) + bd$$
At this point, we have to state that $\sigma(10^k a) = \sigma(a)$ (multiplying $a$ by a power of $10$, we are only adding ''0''s at the end of the number which do not give any contribution to $\sigma$). Hence:
$$\sigma(nm) = \sigma(\sigma(ac) + \sigma(b+d) + \sigma(bd)) = \sigma \left((a \times c) + \sigma(b+d) + (b \times d) \right)$$
In general, if $n = \sum_{i=0}^{+\infty} 10^i n_i$ and $m = \sum_{i=0}^{+\infty} 10^i m_i$, then:
$$\sigma(nm) = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty}10^{i+j}n_i m_j\right) = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty} (n_i \times m_j)\right)$$
In a certain way, I generalize the product $\times$ for any integers $n$ and $M$ as follows:
$$ n \times m = \sigma\left(\sum_{i=0}^{+\infty}\sum_{j=0}^{+\infty} (n_i \times m_j)\right)$$
This is the last structure I'm facing with. Is there some one that ever seen this? Are there some theoretical results?
 A: You are doing multiplication $\pmod 9$, where you only care about the remainder on division by $9$.  Usually one uses $0$ instead of $9$, but that is not important.  You can also add $\pmod 9$, things have additive inverses, and multiplication distributes over addition.  This makes it the ring $\Bbb {Z/9Z}$  This works with any modulus.  If the modulus is prime, you get a field, because all the elements (except $0$) have a multiplicative inverse and you can divide as well.  You might read here
A: Your $\sigma(n)$ is just the remainder when $n$ is divided by $9$ (except that you write "$9$" rather than "$0$" when $n>0$, but that could be considered just a matter of notation, as modulo $9$, we see that $0$ and $9$ are equivalent). So you have results like $\sigma(nm)\equiv\sigma(n)\sigma(m)$ and $\sigma(n+m)\equiv\sigma(n)+\sigma(m)$, where $\equiv$ means congruence modulo $9$ (or equality in the ring $\mathbb{Z}/9\mathbb{Z}$). See modular arithmetic.
In particular, your notation $\times$ seems redundant: your $a\times b$ (or $n \times m$) means nothing other than $\sigma(ab)$ (or $\sigma(nm)$), so your last equation could well be written $\sigma(nm)=\sigma(\sum_{i,j}n_im_j)$ or be written as $n \times m \equiv \sum_{i, j} n_i \times m_j$.
The convolution that you see in the sums above is simply one that comes from the definition of multiplication in terms of digits.
