Normality is also sufficient, also this doesn't depend on the field $k$. This is essentially exercise 6.1.10 in "Cohen-Macaulay rings" by Bruns and Herzog.

Let $S\subset T$ be affine semigroups. Call $S$ a full subsemigroup if $S= T\cap \mathbb ZS$. It is not hard to show that if $S$ is a positive affine semigroup then it is normal if and only if it is isomorphic to a full subsemigroup of $\mathbb N^n$, for some $n\geq 0$.

The claim in question follows from the general fact that if $S$ is full in $T$ then $k[S]$ is a direct $k[S]$-summand of $k[T]$. The proof is straightforward, denote by $W$ the $k$-algebra spanned by elements in $T-S$, then clearly $k[T]=k[S]\oplus W$ as vector spaces. To show that $W$ is in fact a $k[S]$ module, it is enough to check that if $\alpha\in S$ and $\beta\in T-S$ then $\alpha+\beta\in T-S$. If we assume for the sake of contradiction that $\alpha'=\alpha+\beta\in S$, then $\beta=\alpha'-\alpha\in T$ which implies $\beta\in S$ since $S$ is full.