I'll first recall some definitions here for convenience. Given a finite set $S$ in a real vector space $V$, $$\text{Cone}(S)=\{\sum\limits_{u\in S} \lambda_u u\mid \lambda_u\geq 0\}.$$ An affine semigroup $S\subset M$ is called saturated if for all $k\in \mathbb{N}$ and $m\in M$, $km\in S$ implies $m\in S$.
The following is exercise 1.3.4 from Cox, Little, and Schenck's Toric Varieties: Let $\mathcal{A}\subseteq M$ be a finite set. Prove that the semigroup $\mathbb{N}\mathcal{A}$ is saturated in $M$ if and only if $\mathbb{N}\mathcal{A}=\text{Cone}(\mathcal{A})\cap M$. (Hint: Apply (1.2.2) to $\text{Cone}(\mathcal{A})\subset M_{\mathbb{R}})$.
First of all I think the hint might have a typo, as (1.2.2) doesn't seem to have anything to do with this problem. But anyway, here's what I've got so far:
One inclusion is obvious, so we'll do the other. Let $\mathcal{A}=\{m_1,\dots,m_n\}$ and let $s\in \text{Cone}(\mathcal{A})\cap M$. We can write $s=\sum\limits_{i=1}^n a_im_i$ with $a_i\geq 0$. Now write $a_i=\lfloor a_i\rfloor + \delta_i$, so that $s=\sum\limits_{i=1}^n\lfloor a_i\rfloor m_i + \sum\limits_{i=1}^n \delta_i m_i$. Rearranging we get $$s-\sum\limits_{i=1}^n\delta_i m_i=\sum\limits_{i=1}^n \lfloor a_i\rfloor m_i\in\mathbb{N}\mathcal{A}.$$ Now, if we could ensure the $\delta_i$ were rational, then we'd be done. Indeed, we could multiply this equation through by some large natural number to kill all the denominators, then use the saturatedness of $\mathbb{N}\mathcal{A}$ to conclude that $s\in \mathbb{N}\mathcal{A}$. Can such an assumption be made?
Nathan Lowry: in case you still need clarification on this. Expressing $s$ as a linear combination of the $m_i$ is a linear system of equations with the $m_i$ as matrix, $s$ as right-hand side, and the $a_i$ as indeterminates. Whether a linear system (defined over the rationals) has a solution does not depend on whether you look for a rational or a real solution; if a (real) solution exists at all, any natural method you use (Gauss, Cramer, etc) will give you a rational one.