There should be a way to do this purely with playing around with pullback squares in the $\infty$-category of spectra (alternatively, with homotopy pullbacks in a model category presenting it), but I got a bit lost doing that, so I will give an alternative solution, which does require a bit more prior knowledge but is much shorter. I will work in the $\infty$-category $\mathsf{Sp}$ of spectra. You can perform this solution also in the stable homotopy category with its triangulated structure, except for the reduction step from $K$ to $\mathbb{S}$ below, which at the very least requires a not too naive model category for spectra (i.e. one that is a monoidal model category under the tensor product of spectra).
First, note that it does not matter for this question whether $K(\mathcal{C})$ is a spectrum or an $\mathbb{E}_\infty$-group, because the inclusion of $\mathbb{E}_\infty$-groups into spectra preserves homotopy cofibers. We will compute the cofiber in $\mathsf{Sp}$ because it is a stable $\infty$-category, so that you can compute the cofiber by computing the fiber.
I will prove the following:
Claim 1. If $K\in\mathsf{Sp}$ and $\nabla\colon K\oplus K\to K$ is the fold map, then the map $$K\oplus K\oplus K\to\mathrm{cofib}\left((\nabla,\mathrm{id})\colon K\oplus K\to K\oplus K\oplus K\right)$$
is equivalent to the composite
$$
K\oplus K\oplus K\xrightarrow{-1\oplus 1\oplus 1} K\oplus K\oplus K\xrightarrow{\nabla} K.$$
Here, $-1$ is the ''multiplication by $-1$'' map that any spectrum admits (more on that below). Since $\nabla\colon K^{\oplus n}\to K$ serves as an addition map (for instance in the sense that it induces addition on homotopy groups), this statement implies the statement from your question in case $b=a+2$. The general case is proven entirely analogously, but I want to keep notation simple.
We will first assume a weaker claim.
Claim 2. We have $\mathrm{cofib}(\nabla,\mathrm{id})\simeq K$.
The proof will be given at the end. Assuming Claim $2$, we will now reduce Claim $1$ to the case $K\simeq\mathbb{S}$, the sphere spectrum. Indeed, the map $-1\colon K\to K$ and the map $\nabla\colon K\oplus K\to K$ are both induced by $-1\colon\mathbb{S}\to\mathbb{S}$ and $\nabla\colon\mathbb{S}\oplus\mathbb{S}\to\mathbb{S}$ after applying the tensor product $-\otimes K$ of spectra. For the first, this holds by definition, and for the second, this follows from the fact that $-\otimes K$ preserves colimits in $\mathsf{Sp}$. Moreover, the conclusion of Claim $1$ is a statement about cofibers, and since $-\otimes K$ preserves cofibers, it therefore suffices to prove the claim for $K\simeq\mathbb{S}$.
Now, how is $-1\colon\mathbb{S}\to\mathbb{S}$ defined? Well, we know that $-1\colon\mathbb{Z}\to\mathbb{Z}$ exists, and $$\mathsf{Ab}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}\cong\pi_0(\mathbb{S})\simeq[\mathbb{S},\mathbb{S}],$$ so we obtain a map $-1\colon\mathbb{S}\to\mathbb{S}$ unique up to homotopy, that induces multiplication by $-1$ on $\pi_0(\mathbb{S})$. As said earlier, $\nabla\colon\mathbb{S}\oplus\mathbb{S}\to\mathbb{S}$ induces
$$
\pi_0(\mathbb{S})\oplus\pi_0(\mathbb{S})\to\pi_0(\mathbb{S}), (a,b)\mapsto a+b
$$
on homotopy groups. Write $f$ for the map $(\nabla,\mathrm{id})\colon\mathbb{S}\oplus\mathbb{S}\to\mathbb{S}\oplus\mathbb{S}\oplus\mathbb{S})$. We have a long exact sequence
$$
\ldots\to\pi_0(\mathbb{S}\oplus\mathbb{S})\xrightarrow{f_*}\pi_0(\mathbb{S}\oplus\mathbb{S}\oplus\mathbb{S})\to\pi_{0}(\mathrm{cofib}(f))\to\pi_{-1}(\mathbb{S}\oplus\mathbb{S})\to\ldots,
$$
and by Claim 2 and the above facts about homotopy groups we already know that it takes the form
$$
\ldots\to\mathbb{Z}\oplus\mathbb{Z}\xrightarrow{(a,b)\mapsto(a+b,a,b)}\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\xrightarrow{\psi}\mathbb{Z}\to 0,$$
where $\psi$ is yet to be determined. (We used here that $\pi_{-1}(\mathbb{S})=0$, by the way.) But we can now conclude that $\psi$ is the map $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}, (a,b,c)\mapsto -a+b+c$. Since $$[\mathbb{S}\oplus\mathbb{S}\oplus\mathbb{S},\mathbb{S}]\cong[\mathbb{S},\mathbb{S}]\oplus[\mathbb{S},\mathbb{S}]\oplus[\mathbb{S},\mathbb{S}]\cong\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\cong\mathsf{Ab}(\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z},\mathbb{Z}),$$ each map $\mathbb{S}\oplus\mathbb{S}\oplus\mathbb{S}\to\mathbb{S}$ is determined up to homotopy by how it acts on $\pi_0$, and the map from Claim $1$ induces exactly this map. This proves Claim $1$ for $K\simeq\mathbb{S}$, and as said, this proves all of Claim $1$.
Now let us prove Claim $2$. This is formal nonsense about pullbacks (hence why we do it for general $K$). Since $\mathsf{Sp}$ is a stable $\infty$-category, it suffices to show that $\mathrm{fib}(\nabla,\mathrm{id})\simeq\Omega K$. The first ingredient is that the following is a pullback square in $\mathsf{Sp}$:
$$
\require{AMScd}
\begin{CD}
K @>{(1,-1)}>> K\oplus K\\
@VVV @VV{\nabla}V\\
0 @>>> K
\end{CD}
$$
That we need to put $K$ in the top left corner follows via the same reasoning as in my answer here (I just noticed that this was also a question of yours, what a nice continuity), where we moreover use that $\Sigma\Omega K\simeq K$ because $\mathsf{Sp}$ is stable. That the map $K\to K\oplus K$ appearing above is $(1,-1)\colon K\to K\oplus K$ follows from the long exact sequence on homotopy groups, because we know that $\nabla\colon K\to K$ induces addition on homotopy groups.
Now, formal nonsense about limits commuting with limits gives you for any $f\colon A\to C$ and $g\colon B\to D$ a pullback diagram
$$
\require{AMScd}
\begin{CD}
\mathrm{fib}(f)\oplus\mathrm{fib}(g) @>>> 0\\
@V{i_A\oplus i_B}VV @VVV\\
A\oplus B @>{f\oplus g}>> C\oplus D
\end{CD}
$$
where the maps $i_A\colon\mathrm{fib}(f)\to A$ and $i_B\colon\mathrm{fib}(g)\to B$ are the canonical maps from the fiber. Similar formal nonsense gives for any $(h,j)\colon E\to F\oplus F$ a pullback square
$$
\require{AMScd}
\begin{CD}
E\times_{h,F,j}E @>>> F\\
@VVV @VV{\Delta}V\\
E\oplus E @>{h\oplus j}>> F\oplus F
\end{CD}
$$
where $\Delta\colon F\to F\oplus F$ is the diagonal morphism.
Combining these facts, we can write the pullback square
$$
\require{AMScd}
\begin{CD}
\bullet @>>> 0\\
@VVV @VVV\\
K\oplus K @>{(\nabla,\mathrm{id})}>> K\oplus K\oplus K
\end{CD}
$$
as a composite of pullback squares
$$\require{AMScd}\begin{CD}
\bullet @>>> \bullet @>>> 0\\
@VVV @VVV @VVV\\
K\oplus K @>{\Delta}>>(K\oplus K)\oplus(K\oplus K) @>{\nabla\oplus\mathrm{id}}>> K\oplus K\oplus K
\end{CD}$$
and then conclude that the pullback square
$$
\require{AMScd}
\begin{CD}
P @>>> 0\\
@VVV @VVV\\
K@>{(1,-1)}>> K\oplus K
\end{CD}
$$
satisfies that $P\simeq\mathrm{fib}(\nabla,\mathrm{id})$. Now, as a finishing touch, we compute $P$ by considering the following diagram
$$
\require{AMScd}
\begin{CD}
P @>>> 0 @>>> 0\\
@VVV @VVV @VVV\\
K @>{\Delta}>> K\oplus K@>{1\oplus -1}>>K\oplus K
\end{CD}
$$
The right-hand square is a pullback square because $(1,-1)\colon K\oplus K\to K\oplus K$ is an equivalence, and the outer rectangle is also a pullback square by definition of $P$. Therefore the left-hand square is also a pullback square. We already said above how to compute pullbacks along a diagonal morphism: we conclude that $P$ fits into a pullback diagram
$$
\require{AMScd}
\begin{CD}
P @>>> 0\\
@VVV @VVV\\
0 @>>> K
\end{CD}
$$
This means that $\mathrm{fib}(\nabla,\mathrm{id})\simeq P\simeq\Omega K$, as required. This proves Claim $2$, which means we have won.
P.S. You might wonder why we cannot show using our prove of Claim $2$ that the map $K\oplus K\oplus K\to K$ given by the cofiber is the correct one. This would require finding the fiber of $P\to K\oplus K$, hitting that fiber with $\Sigma$, and multiplying the map by $-1$. However, in the proof that $\mathrm{fib}(\nabla,\mathrm{id})\simeq\Omega K$, we juggled a lot of pullback squares around and changed some signs, so it was too confusing for me to actually show we get the correct map $K\oplus K\oplus K\to K$ like this, hence why I reduced to $K\simeq\mathbb{S}$ and passed to homotopy groups. If someone finds a direct proof, we can remove all these homotopy group considerations, and moreover we would get a proof that for more general $\infty$-categories than just that of spectra.
