Suppose we have a numerical monoid $S=\langle g_1,g_2,...,g_n\rangle$. A factorization for an element $n\in S$ is an $n$-tuple $a=(a_1,a_2,...,a_n), a_i\in \mathbb{N}$ satisfying $a_1g_1+a_2g_2+...+a_ng_n=n$. The length of such an $n$-tuple is the value $|a|=a_1+a_2\:+\:...+\:a_n$. Let $Z(n)$ denote the set of all factorizations of an element $n$ in $S$.
The length set of an element $n\in S$, denoted $\mathcal{L}(n)=\{|a|: a\in Z(n)\}$. The length set of the numerical monoid $S$, $\mathcal{L}(S)=\{\mathcal{L}(n):n \in S\}$
The question is, given two distinct numerical monoids $S=\langle g_1,g_2,...,g_n\rangle, S^{\prime}=\langle h_1,h_2,...,h_n\rangle$, when is it true that $L(S)≠L(S^{\prime})$. More specifically, when is it not true that $L(S)=\mathbb{N}=L(S^{\prime})$?
The issue is that you are thinking of the length set of the numerical monoid wrong. It is a set of the lengths sets of elements of the numerical monoid.
So we can have that the length sets of monoid $S$ and $S'$ are not equal if say there is an element $n\in S$ which has factorizations of lengths $\{4, 7, 12\}$, yet no element of $S'$ has that exact same length set. Hence $\{4, 7, 12\} \in L(S)$ but $\{4, 7, 12\} \notin L(S')$.