This is really just a matter of understanding the actual form of the excision isomorphism.
But we can pick $U$, $\sigma$, and $x$ somewhat carefully, to make this easier.
Instead of working with the given $\Delta$-complex structure, let's work with its 2nd barycentric subdivision, which is guaranteed to be an actual simplicial complex whose individual simplices are actually embedded. You probably know that a homology class is invariant under subdivision, so the two classes formed by summing the simplices of the 2nd barycentric subdivision and by summing the 2-simplices of the original $\Delta$-complex structure are equal. That gives us the freedom to work with teh 2nd barycentric subdivision.
Choose $\sigma$ to be a 2-simplex of the 2nd barycentric subdivision. And now choose $U$ to be a regular neighborhod of $\sigma$, chosen so small that $\sigma$ is the unique 2-simplex of the 2nd barycentric subdivision that is contained in $U$; because it's a regular neighborhood of a closed disc in a manifold, $U$ is homeomorphic to $\mathbb R^2$.
Finally, pick $x$ to be a point in the interior of $\sigma$.
What we need to show is that if $c$ is the sum of all simplices of the 2nd barycentric subdivision then the class $[c] \in H_2(M,M-x)$ is the equal to the image, under excision, of the class $[\sigma] \in H_2(U,U-x)$. And this is really just a matter of understanding a concrete description of the excision homomorphism.
Excision can be described like this:
If you have a $k$-cycle $c = \sum a_i \tau_i$ of $(M,M-x)$, and if each $\tau_i$ is contained in either $M-x$ or $U$, and if $c'$ is obtained from $c$ by discarding all terms $a_i\tau_i$ such that $\tau_i$ is contained in $M-x$, then $c'$ is a $k$-cycle of $(U,U-x)$, and the excision map $H_k(U,U-x) \to H_k(M,M-x)$ takes $[c']$ to $[c]$.
Now let's apply this. Certainly the given $c$ is a cycle of $M$, and it is in fact a fundamental cycle, representing the fundamental class $[M]$. So $c$ is certainly also a cycle of $(M,M-x)$. Applying the above description of excision, the terms that we remove from $c$ are all of the terms except for $\sigma$, which is the only term contained in $U$. So all that's left is $c'=\sigma$, and we're done.
Theorem 1: Let $A\subseteq_\text{open} X,\ B\subseteq_\text{open} Y$. Consider the realtive chain complexes $\mathcal C=C_*(X,A)$ and $\mathcal C'=C_*(Y,B)$. Then there is a natural split short exact sequence $$0\to \bigoplus_{p+q=n}H_p(X,A)\otimes H_q(Y,B)\xrightarrow{\text{cross product}} H_n(X\times Y,A\times Y\cup X\times B)\to \bigoplus_{p+q=n-1}\text{Tor}\big(H_p(X,A),H_q(Y,B)\big)\to 0$$
Theorem 2: For a manifold $X$ of dimension $d$ and $x\in X\backslash \partial X$ we have $H_n(X,X\backslash x)=\begin{cases} \Bbb Z &\text{ if }n=d,\\ 0 &\text{ otherwise.}\end{cases}$
Proof: Use excision theorem considering a closed coordinate ball $\overline B\cong \{z\in \Bbb R^d:|z|\leq 1\}$ with $x\in \text{int}(\overline B)$.
Corollary 1: For two manifolds $M,\ \dim M=m$ and $N,\ \dim N=n$ with $x\in M\backslash \partial M,\ y\in N\backslash \partial N$ we have the isomorphism $$H_m(M,M\backslash x)\otimes H_n(N, N\backslash y)\xrightarrow[\cong]{\text{cross product}} H_{m+n}\big(M\times N,M\times N\backslash (x,y)\big)$$
Proof: Combine Theorem 1 and Theorem 2.
Observation 1: For two manifolds $M,\ \dim M=m$ and $N,\ \dim N=n$ with $x\in M\backslash \partial M,\ y\in N\backslash \partial N$ and orientations $\{\mu_p:p\in M\backslash \partial M\},\ \{\nu_q:q\in N\backslash \partial N\}$ we have the following commutative diagram
where as usual $x\in U\subseteq_\text{open}M$ and $\mu_U\in H_m(M, M\backslash U)\cong \Bbb Z$ is a generator such that $\mu_U\longmapsto \mu_p$ for each $p\in U$ under the inclusion induced map $H_m(M,M\backslash U)\to H_m(M,M\backslash p)$. Similarly, define $\nu_V$.
The bottom arrow is an isomorphism due to Corollary 1. A similar reasoning(cross-product argument) says the top arrow is also an isomorphism. The left arrow is an isomorphism due to the choice of $U$ and $V$ while defining orientations. Now, the commutativity of the square implies, the right arrow is an isomorphism.
In other words, $\{\mu_p\times \nu_q: p\in M\backslash \partial M,
q\in N\backslash \partial N\}$ defines an orientation for $M\times N$.
Theorem 3: Let $M,N$ be two compact oriented manifolds with the fundamental classes $[M]\in H_m(M,\partial M),\ [N]\in H_n(N,\partial N)$.
Then, the fundamental class $[M\times N]\in H_{m+n}\big(M\times N, \partial(M\times N)\big)$ is same as the $[M]\times [N]$.
Proof: Consider $x\in M\backslash \partial M, y\in N\backslash \partial N$, then $(x,y)\in M\times N\backslash \partial(M\times N)$. So, the inclusion induce map $H_{m+n}\big(M\times N,\partial(M\times N)\big)\longrightarrow H_{m+n} \big(M \times N, M \times N \backslash (x, y)\big)$ is an isomorphism. Now, consider the commutative diagram below.
The bottom arrow is an isomorphism due to Corollary 1. The left arrow is isomorphism as $x\in \text{int}(M),\ y\in \text{int}(N)$. So, the top arrow is an isomorphism, i.e., the cross product map $H_m (M, \partial M ) \otimes H_n (N, \partial N )\xrightarrow{\times}H_{m+n} \big(M \times N, \partial(M \times N )\big)$ sends $[M]\otimes [N]$ to $[M\times N]$.
Best Answer
The maps $H_n(M;R) \to H_n(M,M-x;R)\cong R$ are not just isomorphisms of groups but of $R$-modules. In particular, they map generators as an $R$-module to generators as an $R$-module (which, in the $R$-module $R$, are the units).