I claim that the proof that singular homology doesn't depend on homotopy type (of the space or of the maps between spaces) is a special case of a combinatorial result. Moreover, the essence of the proof is the same; the proof strategy is the same. So maybe this should be the primary thing and the topological proof is a mere corollary. I haven't read the paper you mention, however I doubt anyone could find this proof unsatisfactory since it almost exactly the simplicial analogue of the usual proof for the topological case that e.g. is found in Hatcher's Algebraic Topology.
We can also extend this to the reduced homology of simplicial sets; see below.
Singular homology of a space $Z$ is nothing but the simplicial homology of $\mathsf{Sing}(Z)$; for a continuous function $f:X\to Y$ the induced maps on singular chain complexes and singular homology are nothing but the induced maps on simplicial chain complexes and homology from $\mathsf{Sing}(f)$; and if the two continuous $f,g:X\to Y$ are homotopic via some $H$ then: $$H’:\mathsf{Sing}(X)\times\Delta^1\overset{1\times\eta}{\longrightarrow}\mathsf{Sing}(X)\times\mathsf{Sing}(|\Delta^1|)\cong\mathsf{Sing}(X\times|\Delta^1|)\overset{\mathsf{Sing}(H)}{\longrightarrow}\mathsf{Sing}(Y)$$Is a simplicial homotopy between $\mathsf{Sing}(f)$ and $\mathsf{Sing}(g)$.
See here. For $n\ge1$ we can simplicially, or combinatorially, "triangulate" the simplicial prism $\Delta^n\times\Delta^1$. For $0\le j\le n$ write $q_j$ for the function: $$[n+1]\to[1]:\,\begin{pmatrix}x:&0&1&\cdots&j&j+1&j+2&\cdots&n+1\\q_j(x):&0&0&\cdots&0&1&1&\cdots&1\end{pmatrix}$$The pair $(\sigma_j,q_j)$ defines an $(n+1)$-simplex of the simplicial prism $\Delta^n\times\Delta^1$; call it $p_j:\Delta^{n+1}\to\Delta^n\times\Delta^1$. It can be shown (without conceptual difficulty but the details are tedious) that $\Delta^n\times\Delta^1$ is the colimit of the following diagram of simplicial sets: $$\Delta^{n+1}\overset{\delta_1}{\longleftarrow}\Delta^n\overset{\delta_1}{\longrightarrow}\Delta^{n+1}\overset{\delta_2}{\longleftarrow}\cdots\overset{\delta_{n-1}}{\longrightarrow}\Delta^{n+1}\overset{\delta_n}{\longleftarrow}\Delta^n\overset{\delta_n}{\longrightarrow}\Delta^{n+1}$$With $(n+1)$ appearances of $\Delta^{n+1}$ and $n$ appearances of an intersection $\Delta^n$. The cocone from this diagram to $\Delta^n\times\Delta^1$ is provided exactly by the maps $p_j:\Delta^{n+1}\to\Delta^n\times\Delta^1$ for $j=0,1,\cdots,n$. That this makes a colimit is conceptually the statement that the $p_\bullet$ "triangulate" the prism, and taking geometric realisations gives the canonical triangulation (and proves that it actually is a triangulation, which isn't obvious and has been asked on this site a few times) of the topological prism $|\Delta^n|\times|\Delta^1|$. None of that is necessary here, but it is background.
To establish the claim about homology we first need to extend to the case of $n=0$; clearly, the "prism" $\Delta^0\times\Delta^1$ is isomorphic to $\Delta^1$ so there is a very obvious triangulation, given just by the map $p_0:\Delta^1\to\Delta^0\times\Delta^1$, $p_0=(\sigma_0,1)=(\sigma_0,q_0)$.
For our setup, we take simplicial sets $X,Y$ and a simplicial homotopy of maps between them, $H:X\times\Delta^1\to Y$, $f\simeq g$. The claim is that the induced maps on homology $f_\ast$ and $g_\ast$ are equal, which is an abstract corollary of the fact that $f,g$'s maps on chain complex are chain homotopic.
Intuitively, given a simplex $x\in X_n$ we have a composite map: $$\Delta^{n+1}\overset{p_j}{\longrightarrow}\Delta^n\times\Delta^1\overset{x\times1}{\longrightarrow}X\times\Delta^1\overset{H}{\longrightarrow}Y$$And letting $j$ vary will let the homotopy 'act' over the whole prism. We have abused simplices with their corresponding maps $\Delta^\bullet\to X$ by the Yoneda lemma; unwinding the Yoneda lemma shows that this map $\Delta^{n+1}\to Y$ describes the $(n+1)$-simplex $H(s_jx,q_j)$ of $Y$.
For $n\ge0$ define the prism operator $P_n:C_n(X)\to C_{n+1}(Y)$ by mapping every $x\in X_n$ to the chain $\sum_{j=0}^n(-1)^j\cdot H(s_jx,q_j)$. For $n<0$ we take $P_n$ to be the zero map.
We expect the chain homotopy relation (or something similar, modulo signs) $\partial P+P\partial=g-f$ to hold because the boundary of the prism is just the top and bottom faces ($g$ and $f$) and the vertical walls; there is one such wall for every face of the $n$-simplex and these walls are nothing but prisms of lower dimension, so we have one prism per boundary face… that is, we have $P\partial$. Loosely speaking, of course.
The chain homotopy condition is trivial in degrees $n\le-1$; in degree zero, we need only check that $g-f\equiv\partial_1P_0$. So fix a vertex $v\in X_0$; we have: $$\begin{align}\partial_1P_0(v)&=d_0 H(s_0v,1)-d_1 H(s_0v,1)\\&=H(d_0s_0v,d_0)-H(d_1s_0v,d_1)\\&=H(v,d_0)-H(v,d_1)\\&=g(v)-f(v)\end{align}$$As desired. The computation for $n\ge1$ is more fiddly.
Observe that $d_0q_0=d_0!$ and $d_{n+1}q_n=d_1!$ where $!$ is the constant function $[n+1]\to[0]$. Other than these fringe cases, we have the general relations: $$d_kq_j=\begin{cases}q_j&k>j\\q_{j-1}&k\le j\end{cases}$$There are also the standard simplicial relations: $$d_ks_j=\begin{cases}1&k=j,j+1\\s_jd_{k-1}&k>j+1\\s_{j-1}d_k&k<j\end{cases}$$
We can then calculate, for $x\in X_n$ and $n\ge1$: $$\begin{align}\partial_{n+1}P_n(x)&=\partial_{n+1}H(s_0x,q_0)+(-1)^n\partial_{n+1}H(s_nx,q_n)+\sum_{j=1}^{n-1}(-1)^j\partial_{n+1}H(s_kx,q_k)\\&=H(x,d_0!)-H(x,d_1!)+\color{red}{H(x,q_{n-1})-H(x,q_0)}\\&+\sum_{k=2}^{n+1}(-1)^kH(s_0d_{k-1}x,q_0)+(-1)^n\sum_{k=0}^{n-1}(-1)^kH(s_{n-1}d_kx,q_{n-1})\\&+\sum_{j=1}^{n-1}\left(\sum_{0\le k<j}(-1)^{k+j}H(s_{j-1}d_kx,q_{j-1})+\sum_{n+1\ge k>j+1}(-1)^{k+j}H(s_jd_{k-1}x,q_j)\right)\\&+\color{red}{\sum_{j=1}^{n-1}(H(x,q_{j-1})-H(x,q_j))}\\&=g(x)-f(x)-\sum_{k=1}^n(-1)^kH(s_0d_kx,q_0)-\sum_{k=0}^{n-1}(-1)^{k+n-1}H(s_{n-1}d_kx,q_{n-1})\\&-H(s_0d_0x,q_0)-\sum_{j=1}^{n-2}\sum_{0\le k\le j}(-1)^{k+j}H(s_jd_kx,q_j)\\&+H(s_{n-1}d_nx,q_{n-1})-\sum_{j=1}^{n-2}\sum_{n\ge k\ge j+1}(-1)^{k+j}H(s_jd_kx,q_j)\\&=g(x)-f(x)-\sum_{k=0}^n(-1)^{k+0}H(s_0d_kx,q_0)-\sum_{k=0}^n(-1)^{k+n-1}H(s_{n-1}x,q_{n-1})\\&-\sum_{j=1}^{n-2}\sum_{k=0}^n(-1)^{k+j}H(s_jd_kx,q_j)\\&=g(x)-f(x)-\sum_{j=0}^{n-1}\sum_{k=0}^n(-1)^{k+j}H(s_jd_kx,q_j)\\&=g(x)-f(x)-P_{n-1}(\partial_nx)\end{align}$$Giving the desired chain homotopy relation: $$\partial_{n+1}P_n+P_{n-1}\partial_n=g-f$$In degree $n\ge1$. The red summands amount to zero because the series is telescoping.
A similar thing works for reduced homology. If $X$ is an augmented simplicial set then by defining $C_{-1}$ to be the free Abelian group on $X_{-1}$ and defining the boundary $\partial_0:C_0\to C_{-1}$ as the linear extension of the augmentation $\epsilon:X_0\to X_{-1}$, it's clear $\partial_0\partial_1=0$ - so with $C_\bullet=0$ in degree $<-1$ we have a chain complex - we could define the augmented homology of $X$ to be the homology of this chain complex. If $X$ is just a simplicial set then we have its trivial augmentation $X^+$ given by $X_{-1}=\ast$ and $\epsilon$ the only possible map; this is functorial in that simplicial maps $f:X\to Y$ yield obvious maps $f:X^+\to Y^+$. We could define the reduced homology of $X$ to be the augmented homology of $X^+$. Clearly the boundary map is now the map $C_0\to\Bbb Z$, $\sum_j n_jv_j\mapsto\sum_j n_j$ and this is the usual boundary map for topological reduced homology.
We can define the prism $P_{-1}:C_{-1}(X)\to C_0(Y)$ just as the zero map and the chain homotopy conditions are trivially satisfied.
