Proving that the set $\pi_n(X,v)$ is a group (Goerss and Jardine Theorem 7.2) In theorem 7.2 of Goerss and Jardine's book Simplicial Homotopy Theory, the authors ask us to show that identity law and inverse law holds for the set $\pi_n(X,v)$. I am unable to prove these statements.
Question about the inverse was asked before, but it has no answers. And I could not find any alternative reference that spells this out in detail.
Any help will be greatly appreciated.

My thoughts:
The group operation is defined as follows:
Given two $n$ simplices $a$ and $b$ (in $X$) one creates an $n+1$ dimensional $n$-horn $(v,v,...v,a,\, ,b)$ in $X$ and fills it with $\omega$ (since $X$ is assumed to be a Kan complex). Then the multiplication of $a$ and $b$ is defined to be the newly filled face by $\omega$, i.e. $[a\cdot b] = [d_n \omega]$.
To prove the identity law, I will have to set $b = v$ and show that $[a]\cdot [v] = [a]$. But this means that for $\omega$ such that $\partial \omega = (v,v,...,a,d_n \omega,v)$, I have to find a homotopy $d_n\omega \to a$. I was thinking that I can precompose some $\triangle^n \times \triangle^1 \to \triangle^{n+1}$ with $\omega$ and obtain the required homotopy. But I am not able to proceed.
For the inverse axiom, the authors ask us to show that the left multiplication map is bijective. I have no idea where to start.
 A: Identity
There is an easier proof of the identity requirement, assuming we know the product is well defined, though your idea also works.
First the easy proof. The other proof is at the end of the post. Take $\omega = s_na$, which has faces $(v,v,\ldots,v,a,a)$ and $\omega=s_{n-1}a$, which has faces $(v,v,\ldots,a,a,v)$ to prove the left and right identity laws.
Inverses
I'll assume that you already know that this multiplication is well-defined (i.e., independent of choice of representatives of homotopy classes and choice of horn filler) and associative (only necessary for left/right inverses to be equal) here.
I think there is actually an easier proof than that suggested by Goerss and Jardine, given these assumptions, so I'll give this proof instead. (Perhaps I'm making a mistake though.)
Left and right multiplication by $[a]$ are both surjective.
For left multiplication, suppose we are given $[a]$ and $[c]$, and we want to find $[b]$ such that $[a]\cdot [b]=[c]$.
Then we can choose a filler $\omega$ for the $\Lambda^{n+1}_{n+1}$
$(v,v,\ldots,v,a,c,-)$. Take $b=d_{n+1}\omega$, and $\omega$ shows that $[a]\cdot[b]=[c]$, assuming multiplication is well-defined.
Essentially the same proof shows right multiplication is also surjective. Thus there are elements $[b]$ and $[b']$ such that $[a]\cdot [b]=[v]=[b']\cdot[a]$, which implies that $[b]=[b']$ (this is where we use associativity), and $[a]$ is invertible.
Explicit Homotopy for identities
Note that this argument has the benefit of not assuming that the product is well defined a priori.
To make your idea work we can do the following to construct an explicit homotopy.
It can be easier to visualize it as building up a $\Delta^n\times \Delta^1$ out of $\omega$ and a bunch of degeneracies.
As a reminder, the $(n+1)$-simplices in $\Delta^n\times \Delta^1$ have vertices
$(0,0),(1,0),\cdots(n,0),(n,1)$, $(0,0),\cdots, (n-1,0),(n-1,1),(n,1)$, and so on.
There are $n+1$ of these, so let's label these $(n+1)$-simplices as $H_0,\ldots,H_n$ in that order.
$H_i$ has vertices $(0,0),\ldots,(n-i,0),(n-i,1),\ldots,(n,1)$, so we have that
$H_i$ and $H_{i+1}$ share a face with vertices $(0,0),\ldots,(n-i-1,0),(n-i,1),\ldots,(n,1)$. Thus we have the compatibility condition: $d_{n-i}H_i=d_{n-i}H_{i+1}$.
Now take
$H_0=s_nd_n\omega$,
$H_{1}=\omega$, and
$H_i = s_{n-i}a$ for $i>1$.
Checking the compatibility condition, we have $d_nH_0=d_ns_nd_n\omega = d_n\omega=d_nH_1$,
$d_{n-1}H_1=d_{n-1}H_1=a=d_{n-1}s_{n-2}a=d_{n-1}H_2$, and for $i>1$,
$d_{n-i}H_i=d_{n-i}s_{n-i}a=a=d_{n-i}s_{n-i-1}a=d_{n-i}H_{i+1}$.
Moreover, $d_{n+1}H_0=d_{n+1}s_nd_n\omega=d_n\omega$, and
$d_0H_n = d_0s_0a=a$.
Thus we have described a homotopy from $d_n\omega$ to $a$. Additionally, it's not hard to verify that the rest of the faces are all $v$, so this is a homotopy that preserves the boundary of the simplex.
The other identity is the same idea, except now we're given that
$\omega$ has boundary $(v,v,\ldots,v,d_n\omega,a)$, so
we take $H_0=\omega$, $H_i=s_{n-i}d_n\omega$ for $i > 0$, and now we get a homotopy from $a$ to $d_n\omega$.
