If $X$ is an Abelian group, then $\ker_X : \mathrm{Cong}(X) \rightarrow \mathrm{Sub}(X)$ is a bijection. Is there a partial converse?

(All monoids are written additively in this question, even the non-commutative ones.)

Given a monoid $X$, write $\mathrm{Sub}(X)$ for the lattice of submonoids of $X$, and write $\mathrm{Cong}(X)$ for the lattice of monoid congruences on $X$. There is a function $\ker_X : \mathrm{Cong}(X) \rightarrow \mathrm{Sub}(X)$ defined as follows; given a congruence relation $R$ on $X$, we define

$$\ker_X(R) = \{x \in X \mid (x,0) \in R\}$$

It is well-known that if $X$ is an Abelian group, then $\ker_X : \mathrm{Cong}(X) \rightarrow \mathrm{Sub}(X)$ is a bijection. An obvious question is whether the converse holds.

The answer is no. For example, consider $X = \{0,1\}$ and define $+$ on $X$ to mean "OR". Explicitly, $x+y = 1$ iff $x=1$ or $y=1$. Then $X$ is not a group, since $1$ has no additive inverse. However, observe that $X$ has two possible quotients, the degenerate quotient $\{\{0,1\}\}$ and the trivial quotient $\{\{0\},\{1\}\}$. And it has two submonoids, namely $\{0,1\}$ and $\{0\}$ respectively. Furthermore, $\ker_X$ is given as follows.

$$\{\{0,1\}\} \mapsto \{0,1\}$$

$$\{\{0\},\{1\}\} \mapsto \{0\}$$

Hence $\ker_X$ is a bijection, despite that $X$ is not a group.

Question. Is there some kind of a partial converse to the statement of interest? Namely that for all commutative monoids $X$, if $X$ is a group, then $\ker_X : \mathrm{Cong}(X) \rightarrow \mathrm{Sub}(X)$ is a bijection.