Yes, this is true. Here's one way to prove it. For $0\leq j<n$, write $s_j^{n-1}:\Delta^n\to\Delta^{n-1}$ for the map induced by the order-preserving surjection $\{0,\dots,n\}\to\{0,\dots,n-1\}$ that maps both $j$ and $j+1$ to $j$. Say that a singular simplex $\Delta^n\to X$ is degenerate if it factors through $s_j^{n-1}$ for some $j$. Note that the boundary of a degenerate simplex is a linear combination of degenerate simplices: all but possibly two of its faces are degenerate (the two exceptions being the faces corresponding to omitting the vertices $j$ and $j+1$), and those two faces cancel out. So if we write $D_n(X)\subseteq C_n(X)$ for the span of the degenerate simplices, $D_\bullet(X)$ is a subcomplex of $C_\bullet(X)$. Write $N_\bullet(X)=C_\bullet(X)/D_\bullet(X)$.
Now note that given an order-preserving simplicial map $f:X\to Y$, the induced maps $C^\Delta_\bullet(X)\to C^\Delta_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)$ and $C^\Delta_\bullet(X)\to C_\bullet(X)\to C_\bullet(Y)\to N_\bullet(Y)$ are equal, since the $n$-simplices $\sigma$ that you're worried about have the property that $f\circ c_\sigma$ is degenerate. So to show that $f^\Delta_{\bullet_*}=f_{\bullet_*}$, it suffices to show that the map $C_\bullet(Y)\to N_\bullet(Y)$ induces isomorphisms on homology. By the long exact sequence in homology associated to the short exact sequence $0\to D_\bullet(Y)\to C_\bullet(Y)\to N_\bullet(Y)\to 0$, it suffices to show that $D_\bullet(Y)$ has trivial homology.
We can show this by constructing a chain homotopy. Given a degenerate $n$-simplex $\sigma:\Delta^n\to Y$, let $j(\sigma)\in\{0,\dots,n-1\}$ be the least $j$ such that $\sigma$ factors through $s_j^{n-1}$. Now define $H:D_n(Y)\to D_{n+1}(Y)$ by $H(\sigma)=(-1)^{j(\sigma)}\sigma\circ s^n_{j(\sigma)}$. An elementary computation then shows that $H\partial+\partial H:D_n(Y)\to D_n(Y)$ is the identity map. It follows that $D_\bullet(Y)$ has no homology.
To give a sense of the computation $H\partial+\partial H=1$, let's show what happens when you take a $3$-simplex $\sigma$ with $j(\sigma)=1$; the general case works very similarly. Let's write $\sigma=[a,b,b,c]$; all the simplices built from $\sigma$ will be written using similar expressions with the obvious meaning (here $a$, $b$, and $c$ are the vertices of $\sigma$, with $b$ repeated since it factors through $s^2_1$). We then have $H(\sigma)=-[a,b,b,b,c]$, so $$\begin{align}\partial H(\sigma)=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,c]+[a,b,b,c]-[a,b,b,b]\\=&-[b,b,b,c]+[a,b,b,c]-[a,b,b,b].\end{align}$$
On the other hand, $\partial\sigma=[b,b,c]-[a,b,c]+[a,b,c]-[a,b,b]=[b,b,c]-[a,b,b]$. We have $H([b,b,c])=[b,b,b,c]$ and $H([a,b,b])=-[a,b,b,b]$. Thus we get $$H(\partial\sigma)=[b,b,b,c]+[a,b,b,b].$$ When we add together $\partial H(\sigma)$ and $H(\partial\sigma)$, all the terms cancel except $[a,b,b,c]$, which is just $\sigma$.
Although the statements in your post are all true, none of them are correct interpretations of this passage from Hatcher's book.
Instead, what Hatcher is saying is that for any topological space $X$ there exists a simplicial complex $Y$ such that the singular chain complex $S(X)$ and the simplicial chain complex $\Delta(Y)$ are isomorphic as chain complexes. And then, because of this chain complex isomorphism, it follows that the singular homology of $X$ is isomorphic to the simplicial homology of $Y$.
Furthermore Hatcher's description of the isomorphism is defined by bijections, namely one bijection for each $n \ge 0$, between the basis of $S_n(X)$ (i.e. the set of singular $n$-simplices in the topological space $X$) and the basis of $\Delta_n(Y)$ (i.e. the set of $n$ simplices of the simplicial complex $Y$).
Best Answer
Having the same homology and "inheriting" $\Delta$-structure are very vaguely related things.
A simplest counterexample would be a contractible space $Y$ which is not a simplicial/$\Delta$/CW-complex. For example the comb space. Because it is not locally connected.
Homotopy equivalences rarely preserve topological properties.