Will this construction of $K(M)$ work better for you? I assume that $M$ is a commutative semigroup; that is, $M$ does not necessarily have the zero element. In a sense, $K(M)$ is a generalization of how we construct $\mathbb{Z}$ from the natural numbers, or $\mathbb{Q}_{\neq 0}$ from the nonzero integers.
Equip $M\times M$ with the following equivalence relation $\sim$ defined as follows: for $x,y,z,w\in M$, $$(x,y)\sim (z,w)\text{ if and only if there exists }m\in M\text{ such that }x+w+m=z+y+m\,.$$
Let $G:=(M\times M)/\sim$ be the set of equivalence classes of $M\times M$ with respect to $\sim$. Note that $G$ is an abelian group with addition $+$ defined by
$$\big[(x,y)\big]_G+\big[(z,w)\big]_G:=\big[(x+z,y+w)\big]_G\text{ for all }x,y,z,w\in M\,,$$
where $\big[(x,y)\big]_G$ denote the equivalence class in $G$ that contain $(x,y)\in M\times M$. The zero element of $G$ is just the class
$$0_G:=\big[(x,x)\big]_G\,(\text{for any }x\in M)\,.$$
For each $x,y\in M$, the inverse of $\big[(x,y)\big]_G$ in $G$ is simply $\big[(y,x)\big]_G$. Then, $K(M)$ is isomorphic to $G$, and the map $\gamma:M\to K(M)$ is precisely the map sending $x\in M$ to
$$\gamma(x):=\big[(x+y,y)\big]_G\,(\text{for any }y\in M)\,.$$
This map is injective if and only if $M$ has the cancellative property (i.e., for $x,y,z\in M$, $x+y=x+z$ implies $y=z$). The map is bijective if and only if $M$ is an abelian group.
To show that $G$ indeed satisfies the universal property, let $f:M\to A$ be a semigroup homomorphism from $M$ to an abelian group $A$. Define $f_*:G\to A$ via
$$\big[(x,y)\big]_G\mapsto f(x)-f(y)$$
for all $x,y\in M$. Show that $f_*$ is a group homomorphism, and that $f_*\circ \gamma=f$. If $g:G\to A$ is another group homomorphism such that $g\circ \gamma=f$, then
$$g\Big(\big[(x+y,y)\big]_G\Big)=(g\circ\gamma)(x)=f(x)\,,$$
for any $x,y\in M$. That is, for all $x,y,z\in M$, we have
$$\begin{align}g\Big(\big[(x,y)\big]_G\Big)+f(y)&=g\Big(\big[(x,y)\big]_G\Big)+g\Big(\big[(y+z,z)\big]_G\Big)\\&=g\Big(\big[(x+y+z,y+z)\big]_G\Big)=f(x)\,.\end{align}$$
This proves that
$$g\Big(\big[(x,y)\big]_G\Big)=f(x)-f(y)=f_*\Big(\big[(x,y)\big]_G\Big)$$
for $x,y\in M$, implying that $g=f_*$.
To make your visualization of $K(M)$ a bit stronger, here are examples.
Consider $M:=(\mathbb{Z},\max)$ (that is $x+y$ is defined to be $\max\{x,y\}$ here). Then, show that $K(M)$ is the trivial abelian group $0$.
Consider $M:=(\mathbb{Z}_{>0},\gcd)$ (that is, $x+y$ is defined to be $\gcd(x,y)$ here). Then, show that $K(M)$ is the trivial abelian group $0$.
Consider $M:=(\mathbb{Z}_{\geq 0},+)$. Then, show that $K(M)\cong(\mathbb{Z},+)$.
Consider $M:=(\mathbb{C}_{\neq 0},*)$, where $x*y$ is defined to be $|xy|$. Prove that $K(M)\cong (\mathbb{R}_{>0},\cdot)$.
Consider $M:=(\mathbb{Z}_{>0},\cdot)$. Show that $K(M)\cong(\mathbb{Q}_{>0},\cdot)$.
