A description is given by:
Proposition. Let $L$ be a nonempty $n$-ary subsemigroup of $\mathbf{N}$, and generating $\mathbf{Z}$ as a subgroup. Then there exists $k\ge 1$ dividing $n-1$ and $0<\ell<k$ coprime to $k$ such that $L$ is a cofinite subset of the arithmetic progression $k\mathbf{N}+\ell$.
(The condition that $L$ generates $\mathbf{Z}$ is not serious: in general $L$ generates a subgroup of $r\mathbf{Z}$ of $\mathbf{Z}$ and we deduce that $L$ is a cofinite subset in $r(k\mathbf{N}+\ell)$ for some $k$ dividing $n-1$ and $0<\ell<k$ coprime to $k$.)
Lemma 1 Let $L$ be an $n$-ary subsemigroup of $\mathbf{N}$. Then $L$ is ultimately periodic: there exists $k$ such that the symmetric difference of $k$ and $L+k$ is finite, and the projection of $L$ modulo $k$ is an $n$-ary subsemigroup of $\mathbf{Z}/k\mathbf{Z}$. Moreover, if $k$ is chosen minimal and $L$ generates $\mathbf{Z}$ as a subgroup, then $k$ divides $n-1$.
Hence, the classification "modulo finite subsets" reduces to the classification of $n$-ary subsemigroups of $\mathbf{Z}/(n-1)\mathbf{Z}$.
Proof of Lemma 1. If $L$ is empty the result is clear; assume otherwise. Choose $k_0$ in $L$. Then $L+(n-1)k_0\subset L$. Hence the sequence indexed by $m$ of projections modulo $n-1$ of $L\cap\{m(n-1)+1,m(n-1)+2,\dots,m(n-1)+n-1\}$ is an ascending sequence of subsets of $\mathbf{Z}/(n-1)\mathbf{Z}$ and hence stabilizes. So $L$ is eventually $(n-1)$-periodic. If $k,k'$ are eventual periods, then clearly so is $\gcd(k,k')$. Let $k$ be the smallest eventual period (so $k$ divides $n-1$). For large enough $m$, the subset $L\cap\{m(n-1)+1,m(n-1)+2,\dots,m(n-1)+n-1\}$ doesn't depend on $m$ and is a $n$-ary subsemigroup $L_k$ of $\mathbf{Z}/k\mathbf{Z}$. Moreover if $L$ generates $\mathbf{Z}$, the $L_k$ generates $\mathbf{Z}/k\mathbf{Z}$. (Indeed otherwise it would mean that for some prime $p$, $L\smallsetminus p\mathbf{Z}$ is finite. But choosing $n-1$ large elements in $L\cap p\mathbf{Z}$ and one element in $L\smallsetminus p\mathbf{Z}$ yields a large element in $L\smallsetminus p\mathbf{Z}$, contradiction.)
So for each $n$ and $k$ dividing $n-1$, we are reduced to finding the non-empty $n$-ary subsemigroups of $\mathbf{Z}/k\mathbf{Z}$ that generate $\mathbf{Z}/k\mathbf{Z}$ (call this $n$-templates), which have no smaller period (i.e. for no strict divisor $\ell$ of $k$ are inverse image of an $n$-template in the quotient $\mathbf{Z}/\ell \mathbf{Z}$.
Lemma 2. Every $n$-template in $\mathbf{Z}/k\mathbf{Z}$ is reduced to a singleton $\{\ell\}$ with $\ell$ coprime to $k$.
Proof: endow $\mathbf{Z}/k\mathbf{Z}$ with the obvious distance ($d(\ell,\ell')$ is the minimum absolute value of a representative of $\ell-\ell'$). Suppose by contradiction that a template $T$ is not reduced to a singleton and let $a,b$ be at minimal distance, say $k'$ (so $k'\le k/2$). Considering all possible $n$-ary sums of $a$ and $b$, we see that $T$ contains the arithmetic progression of period $b-a$ $\{a,b,2b-a,3b-2a,\dots ,2a-b\}$. Since $k$ is the minimal period for $T$, $T$ cannot be $(b-a)$-periodic and hence there is another element in $T$. But this contradicts the minimality of the distance.
Proof of the proposition: Combining the lemmas shows that $L$ eventually coincides with $k\mathbf{N}+\ell$ for some $k$ dividing $n-1$ and $\ell\in\{0,\dots,k-1\}$ coprime to $k$ (I use the convention $0\in\mathbf{N}$). It remains to show that $L\subset k\mathbf{N}+\ell$. Indeed otherwise there's in $L$ an element of the form $km+\ell'$ with $\ell'\neq \ell$ mod $k$. Taking with $n$-ary sum with $n-1$ large elements in $L\cap (k\mathbf{N}+\ell)$ yields a contradiction.
Examples:
$n$ arbitrary, $k=1$: the only $n$-template in $\mathbf{Z}/1\mathbf{Z}$ is $\mathbf{Z}/1\mathbf{Z}$. This corresponds to the case when $L$ is cofinite. For $n=2$, $k$ divides $n-1=1$ so there is no other case.
$n=2$: here $n-1=1$ so $k=1$ and the only $2$-template in $\mathbf{Z}/1\mathbf{Z}$ is $\mathbf{Z}/1\mathbf{Z}$. It means $L$ is a cofinite subset (classical case).
For $k\neq 1$:
$n=3$: the only nonempty 3-template in $\mathbf{Z}/2\mathbf{Z}$ is $\{1\}$. It corresponds to the case when $L$ eventually consists of all odd numbers.
$n=4$: the only nonempty 4-templates in $\mathbf{Z}/3\mathbf{Z}$ are $\{1\}$ and $\{2\}$. They correspond to the case when $L$ eventually consists of all numbers equal to 1, resp. 2, modulo 3.
$n=5$: either $k=2$ and then $L$ eventually consists of all odd numbers, or $k=4$ and then $L$ eventually consists of all numbers that equal $i$ modulo $4$ (for some fixed $i=1$ or $i=3$).
Etc.
