The key point is that the inclusion of $X$ into the mapping cylinder is a cofibration. It is a general fact that if $A \hookrightarrow B$ is a cofibration and a homotopy equivalence, then $A$ is a deformation retract of $B$. One way to think of this is to use the model structure on topological spaces (due to Strom) where (closed) Hurewicz cofibrations are cofibrations, Hurewicz fibrations are fibrations, and weak equivalences are homotopy equivalences.
So let's prove the following more general fact. Let $A \hookrightarrow B$ be an acyclic cofibration in a (closed) model category, where $B$ is fibrant. Then $B$ "deformation retracts" onto $A$. Let us assume that our model categories have functorial factorizations, so there is always a functorial cylinder object $A \times I$ for $A$.
To see this, let us first show that $B $ retracts onto $A$. By replacing the model category with the model category of objects under $A$, we can assume that $A$ is the initial object $\emptyset$, and $B$ is cofibrant in such a way that $\emptyset \to B$ is a weak equivalence.
This we can do by considering the lifting diagram with $\emptyset \to B$ and $\emptyset \to \ast$ (where $\ast$ is the final object).
We get a retraction $B \to \emptyset$.
Now we want a "deformation retraction." Let us consider the two maps $B \rightrightarrows B$ given by the identity and the retraction. We want a homotopy between the two.
But we can consider the lifting diagram with $B \sqcup B \to B \times I$ (for $B \times I$ a cylinder object) and $B \to \ast$. Since $\emptyset \to B$ is an acyclic cofibration, so is $\emptyset \to B \sqcup B$, and two-out-of-three shows that $\emptyset \to B \times I$ is a weak equivalence too; thus $B \sqcup B \sqcup \emptyset \times I \to B \times I$ is a trivial cofibration. Thus a lifting exists in the diagram, which is the map $B \times I \to B$ that we wanted.
Regarding 1:
Consider a homomorphism $f: \mathbb{Z} \ast \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$.
If $a$ and $b$ are the generators of $\mathbb{Z} \ast \mathbb{Z}$ then consider what $ab$ and $ba$ map to:
$f(ab) = f(a)f(b) = f(b) f(a) = f(ba)$
Where the second equality holds because $\mathbb{Z} \times \mathbb{Z}$ is commutative. But $ab \neq ba$, hence $f$ is not injective.
Regarding 2:
Consider any $(n, m) \in \mathbb{Z} \times \mathbb{Z}$. Then note that $i_\ast (a^n b^m ) = i_\ast (a^n) i_\ast (b^m) = n i_\ast (a) m i_\ast (b) = n (1,0) + m (0,1) = (n,m)$. So $i_\ast$ is surjective.
Regarding 3:
There is a theorem as follows (Hatcher, page 36):
If a space $X$ retracts onto a subspace $A$, then the homomorphism $i_\ast : \pi_1(A, x_0)\to \pi_1(X, x_0)$ induced by the inclusion $i : A \hookrightarrow X$ is injective. If $A$ is a deformation retract of $X$ , then $i_\ast$ is an isomorphism.
In 1 you showed that $i_\ast$ is not injective hence (by contraposition) you get that $S^1 \times S^1$ does not retract onto $S^1 \vee S^1$.
Regarding 4:
There is another theorem (Hatcher page 46):
If $\varphi : X \to Y$ is a homotopy equivalence, then the induced homomorphism $\varphi_\ast :\pi_1(X,x_0)\to \pi_1 (Y,\varphi(x_0))$ is an isomorphism for all $x \in X$.
You know that the induced homomorphism isn't injective hence it cannot be an isomorphism and hence $i_\ast$ cannot be a homotopy equivalence.
Best Answer
Let $r\colon I\times I\setminus\{p\}\to \partial (I\times I)$ be the retraction you mentioned. Can you show that there exists a retraction $\tilde{r}\colon (I\times I\setminus\{p\})/{\sim}\to \partial (I\times I)/{\sim}$ which make the following square commute?
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} I\times I\setminus\{p\} & \ra{r} & \partial (I\times I) \\ \da{q} & & \da{q|_{\partial(I\times I)}} \\ (I\times I\setminus\{p\})/{\sim} & \ras{\tilde{r}} & \partial (I\times I)/{\sim} \\ \end{array} $$