As Noel Lundström pointed out in his comment, the square in the first diagram is not a pushout diagram because the homotopy extension property does not require that $\tilde H$ is unique. If you require uniqueness, then for example $\{0\} \hookrightarrow I$ would be not a cofibration. By the way, in the bottom right corner of a pushout square you would not find $X \times I$, but the adjunction space $X \cup_{i_0} A \times I$ which can be identified with $X \times \{0\} \cup A \times I$ if $A$ is closed in $X$.
You correctly invoke the "adjunction isomorphism" $\phi : \hom(X \times I,Y) \to \hom(X,Y^I)$ which is known as the exponential law. There mere existence of some natural bijection $\phi$ is not sufficient for our purposes, we need the fact that $\big(\phi(G)(x)\big)(t) = G(x,t)$. This implies the essential formula
$$p_0 \circ \phi(G) = G \circ i_0$$
because $(p_0 \circ \phi(G))(x) = p_0(\phi(G)(x)) = \big(\phi(G)(x)\big)(0) = G(x,0) = (G \circ i_0)(x)$.
The maps $H$ in the first diagram and $h$ in the second diagram are related by $\phi(H) = h$, and so will be the fillers.
Let $\tilde H$ be a filler in the first diagram. Define $\tilde h = \phi(\tilde H) : X \to Y^I$. Since $\tilde H \circ (i \times id_I) = H$, we get $\tilde h \circ i = \phi(\tilde H) \circ i = \phi(\tilde H \circ (i \times id_I)) = \phi(H) = h$ by naturality. Moreover, $p_0 \circ \tilde h = p_0 \circ \phi(\tilde H) = \tilde H \circ i_0 = f$. Hence $\tilde h$ is a filler for the second diagram.
Let $\tilde h$ be a filler in the second diagram. Define $\tilde H = \phi^{-1}(\tilde h) : X \times I \to Y$. Then $\phi(\tilde H \circ (i \times id_I)) = \phi(\tilde H) \circ i = \tilde h \circ i = h = \phi(H)$ by naturality and we conclude $\tilde H \circ (i \times id_I) = H$. Moreover, $\tilde H \circ i_0 = \phi^{-1}(\tilde h) \circ i_0 = p_0 \circ \phi(\phi^{-1}(\tilde h)) = p_0 \circ \tilde h = f$. Hence $\tilde H$ is a filler for the first diagram.
I think Tyrone's comments have clarified that "$H_1$ is a homeomorphism onto its image" means that $H'_1 : A \stackrel{H_1}{\to} H_1(A)$ is a homeomorphism which is obvious by definition of $H$ (note that $H_1(A)$ is the image of $A \times \{1\}$ under the quotient map ($A \times I) + B \to X$). Since certainly $H_1 = G_1 \circ i$, we get
$H_1(A) = G_1(i(A))$ so that $G_1$ restricts to $G'_1 : i(A) \stackrel{G_1}{\to} H_1(A)$. By construction we have $G'_1 \circ i = H'_1$.
Let $i' : A \stackrel{i}{\to} i(A)$ (which is a continuous surjection) and $\phi = (H'_1)^{-1} \circ G'_1$. Then
$$\phi \circ i' = (H'_1)^{-1} \circ G'_1 \circ i' = id_A .$$
This shows that $i'$ must be injective. Thus $i'$ is a continuous bijection and $\phi$ is its inverse which is continuous. Therefore $i'$ is a homeomorphism which means that $i$ is an embedding.
Note that an alternative proof can be based on $X = A \times I / A \times \{0\}$ which is a variant of the cone on $A$. Let $H : A \times I \to X$ be the quotient map and $G_0 : B \to X$ be the constant map $G_0(b) \equiv *$, where $*$ is the equivalence class of $A \times \{0\}$. Now argue as above.
Edited:
Let us try to understand why the mapping cylinder $X$ occurs in the above proof. What follows is perhaps a little more complicated, but I hope it makes it more transparent.
For a space $Z$ let $i^Z_t : Z \to Z \times I, i^Z_t(z) = (z,t)$. This is an embedding for each $t \in I$. The mapping cylinder $X = M(i) = \left((A \times I) + B \right)/(a,0) \sim i(a)$ is the pushout of the pair of maps $i^A_0 : A \to A \times I$ and $i : A \to B$. It comes along with maps $H : A \times I \to X$ and $G_0 : B \to X$ such that $G_0 \circ i = H \circ i^A_0$ which satisfy the universal property of the pushout. These maps are those occurring in your diagram. They are the restrictions of the quotient map $q : (A \times I) + B \to X$ to $A \times I$ and to $B$. Since $(i \times id_I) \circ i^A_0 = i^B_0 \circ i$, there exists a unique map $F : X \to B \times I$ such that $F \circ H = i \times id_I$ and $F \circ G_0 = i^B_0$. Explicitly it is given by $F([a,t]) = (i(a),t)$ and $F([b]) = b$.
Since $i$ is a cofibration, we moreover find a (non-unique) map $G : B\times I \to X$ as in your diagram. By the universal property of the pushout we have $G \circ F = id_X$ because $(G \circ F) \circ G_0 = G \circ i^B_0 = G_0 =id_X \circ G_0$ and $(G \circ F) \circ H = G \circ (i \times id_I) = H =id_X \circ H$. Thus $F$ is an embedding. In fact, each map $e : Y \to Z$ which has a left inverse $r : Z \to Y$ (which means $r \circ e = id_Y$) is an embedding: Clearly $e$ must be injective so that the map $e' : Y \stackrel{e}{\to} e(Y)$ is a continuous bijection with $(e')^{-1} = r\mid_{e(Y)}$ which is continuous.
The map $j_1 : A \stackrel{H \circ i^A_1}{\to} A' = H(A \times \{1\}) \subset X$ is a homeomorphism. We have $j_1(a) = [a,1]$. Trivially $k_1 : i(A) \to A'' = i(A) \times \{1\} \subset B \times I$ is a homeomorphism. Since $F$ is an embedding and $F(A') = F(H(A\times\{1\}) = (i \times id_I)(A\times\{1\}) = i(A) \times \{1\} = A''$, we see that $F' : A' \stackrel{F}{\to} A''$ is a homeomorphism. But for $i' : A \stackrel{i}{\to} i(A)$ we have $F' \circ j_1 = k_1 \circ i'$ which implies that $i'$ is a homeomorphism.
Best Answer
Here's another more direct approach. We'll use the following proposition:
Let $f: A \to B$ be a map. TFAE:
We use property 3. Let $W = X \cup_A B$. We want to show that $g^*:[W,Z] \to [X,Z]$ is a bijection for all $Z$. I'll just sketch the idea of surjectivity (and injectivity is very similar).
Let $z: X \to Z$ be a map. Consider $z \circ i: A \to Z$. Since $f$ is a homotopy equivalence, there exists a map $z': B \to Z$ such that $z' \circ f \sim z \circ i$. Since $i$ is a cofibration, there exists a solution $\tilde{h}: X \to Z^I$ to the homotopy extension problem
\begin{CD} A @>{h}>> Z^I\\ @VVV @VV{ev_0}V \\ X @>>> Z \end{CD}
Use this to define the map $w: W \to V$ such that $w \circ g \sim v$. Namely, consider a homotopy $A \times I \to Z$ and create a map $z'' \sim z$ such that $z' \circ f \sim z'' \circ i$. Define $w$ in terms of $z'$ and $z''$ and this gives the appropriate map [details left to you].
Injectivity is a similar idea.