Definition from P.A.Grillet, "Abstract Algebra" (quoted with modifications):
The free commutative monoid on a finite set $X=\{x_1, x_2,..., x_n \}$ is the semigroup of all monomials $x^{a_1}_1 x^{a_2}_2 ... x^{a_n}_n$ (with $a_i \geq 0, a_1 + a_2 + ... + a_n \geq 0)$; the free commutative semigroup on $X = \{x_1,x_2,...,x_n \}$ is the semigroup of all monomials $x^{a_1}_1 x^{a_2}_2 ... x^{a_n}_n$ (with $a_i \geq 0, a_1 + a_2 + ... + a_n > 0)$
The author restricts the definition only to the finite case. What is an infinite case then? I don't see why it would differ very much. To me it looks just like the same, except $\forall x^{a_1}_1 x^{a_2}_2 ... x^{a_i}_i ... \in C \ \ \exists$ only finitely many $a_i \neq 0$. Am I right or no?
Yes, you are essentially right (but including the "only finitely many of the $a_i$ are $\neq 0$" condition is important, so I wouldn't say that it looks "just like the same"!). To be really precise, you also have to change your notation, because the elements of an infinite set cannot always be labelled by $1, 2, 3, \ldots$. So, instead of monomials $x_1^{a_1} x_2^{a_2} x_3^{a_3} \cdots$ corresponding to sequences $\left(a_1, a_2, a_3, \ldots\right) \in \mathbb{N}^{\left\{1,2,3,\ldots\right\}}$, you now want to consider monomials $\prod\limits_{x \in X} x^{a_x}$ corresponding to families $\left(a_x\right)_{x \in X} \in \mathbb{N}^X$.
I think the author just did not want to bother with these technicalities, which is why they restricted themselves to the finite cases.