$\pi_{X}\circ i$ may not be a quotient map( what you call an "identification"). For example, consider the spaces $$X=[0,1]\cup[2,3],\ Y=\{q|\ q\text{ is a rational in } [0,1]\}\cup \{r|\ r \text{ is an irrational in }[2,3]\},$$
and let $x\sim y$ if $|x-y|=0\text{ or }2$. Clearly $Y$ intersects every equivalence class. The set $A=\{[x]|x\in
\mathbb{Q}\cap X\}$ is not open in the space $X/\sim$, for $\pi_{X}^{-1}(A)=X\cap\mathbb{Q}$ is not open in $A$. But $(\pi_{X}\circ i)^{-1}(A)=[0,1]\cap\mathbb{Q}$ is open in $Y$. Thus $\pi_{X}\circ i$ is not a quotient map.
Here's where you made a mistake: the line where you said "$\pi_{X}\circ i$ is the composition of an identification, $\pi_{X}$, and a homeomorphism, hence it is an identification." $i$ is in general not a homeomorphism unless $X=Y$. (If $Y\neq X$, $i$ is not surjective.)
Let $Y$ be any simplicial set and assume we have two maps $f,g : \displaystyle\coprod_{x: \Delta^n \to X}\Delta^n \to Y$ that make the above diagram commute, that is $f\circ (id, j) = g\circ (id, i)$ .
I will use the following convention : an element of the coproduct $\coprod_{i\in I}X_i$ is denoted $(x,i)$, where $x\in X_i$
Now assume $h: X\to Y$ makes the whole thing commute. Let $x\in X_n$ be an $n$-simplex. Then $x$ corresponds to some $\tilde{x} : \Delta^n \to X$ (by the Yoneda lemma), and so $h_n(x) = f((id_{[n]},\tilde{x}))(=g((id_{[n]},\tilde{x}))$ by commutation of the diagram). Therefore if $h$ exists it is unique.
Now define $h$ degree-wise as above, with $f$ : it is well-defined on each degree by the Yoneda lemma. We must show that it is a simplicial map and that it makes the diagram commute, after this by the uniqueness above we will be done. That it makes the diagram commute is quite obvious, as $(id,i),(id,j)$ are surjective, so since $f\circ (id,j) = g\circ (id,i)$ and $h$ is defined through $f$, so this is clear.
Let's now prove that it is a simplicial map. Let $\varphi : [m]\to [n]$ be nondecreasing and $x\in X_n$. Let me write the action of $\varphi$ on the right, as $X$ is a contravariant functor of $[k]$.
Then $h(x\cdot \varphi) = f((id_{[n]},\widetilde{x\cdot\varphi})$, but $\widetilde{x\cdot \varphi}$ is nothing but $\tilde{x}\circ \overline{\varphi}$ where $\overline{\varphi}$ is the induced map $\Delta^m\to \Delta^n$; for this you have to explicit the Yoneda isomorphism.
Also, $(id_{[n]},\tilde{x}\circ \overline{\varphi}) = (id,j)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$ where $y = x\cdot \varphi$, therefore $h(x\cdot \varphi) = g((id,i)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$.
Now $i(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x}) = (\overline\varphi(id_{[m]}), \tilde{x})$ by definition.
Now if you recall the definition of the induced map $\Delta^m\to \Delta^n$ and of the simplicial structure on $\Delta^n$, you see that $\overline\varphi (id_{[m]}) = id_{[n]}\cdot \varphi$. Therefore $h(x\cdot \varphi) =g((id_{[n]}, \tilde{x})\cdot \varphi) = g((id_{[n]},\tilde{x}))\cdot \varphi = h(x)\cdot \varphi$ : $h$ is simplicial; and we are done.
Best Answer
Have a look at the catgorical definition of pushout. It is enough to show that $(V\times W)^\infty$ has the universal property. That's because pushouts are unique up to isomorphism.
I will use $V^\infty\vee W^\infty$, i.e. the wedge sum instead of $(V^\infty\times\{\infty\})\cup(\{\infty\}\times W^\infty)$ because thats what it is.
So we have a commutative diagram
$$\begin{CD} V^{\infty}\vee W^{\infty} @>{i}>> V^{\infty}\times W^{\infty}\\ @VVV @VV{f}V\\ \{\infty\} @>{g}>> (V\times W)^\infty \end{CD} $$
So $i$ is the inclusion, $g(\infty)=\infty$ and
$$f(v,w)=\left\{\begin{matrix}\infty &\text{if }v=\infty\text{ or }w=\infty \\ (v,w) &\text{otherwise}\end{matrix}\right.$$
So this is the first part of the universal property. I encourage you to check that the diagram commutes and that $f$ is continuous.
Lets have a look at the second part of the universal property. Assume that $$\begin{CD} V^{\infty}\vee W^{\infty} @>{i}>> V^{\infty}\times W^{\infty}\\ @VVV @VV{a}V\\ \{\infty\} @>{b}>> P \end{CD} $$
is some commutative diagram. Define
$$\phi:(V\times W)^\infty\to P$$ $$\phi(v,w)=a(v,w)$$ $$\phi(\infty)=b(\infty)$$
So obviously $\phi\circ f=a$ and $\phi\circ g=b$. So all that is left is to show that $\phi$ is continuous. Can you complete the proof?
Side note: If you look carefuly at each function I've defined you will see that there's no magic here. These are just natural choices.