Recall that an elliptic curve over a field k i.e a proper smooth connected curve of genus 1 eq

Prelude
Throughout, I will refer to the distinguished point of $E$ as $\mathrm{\theta <!--\; \theta \; -->}$ instead of 0.
It is important to understand that your rational map p is simply not defined at $\mathrm{\theta <!--\; \theta \; -->}$, which is what you likely mean by saying that the morphism $\mathrm{\phi <!--\; \phi \; -->}:E\to <!--\; \to \; -->{\mathbb{P}}^2$ can "informally" be written as follows:
$P\mapsto <!--\; \mapsto \; -->[x(P):y(P):1]$
I personally think of this scenario in two ways:
1. A continuous extension of this rational map to $\mathrm{\theta <!--\; \theta \; -->}$. This is a more geometric approach and my below explanation is not rigorous (due to the field being completely general here), but I think it is helpful.
2. For a rigorous approach, you have to use a different rational representation of $\mathrm{\phi <!--\; \phi \; -->}$ around $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->E$ and prove that it:
3. has the desired property and
4. is compatible with the one that you use everywhere else on the curve.
Geometric Intuition
Let us think of $k$ as a continuous and friendly field, like $\mathbb{C}$. Let's imagine $E$ as being embedded into some projective space $\mathbb{P}(V)$. For a point $p\in <!--\; \in \; -->V$, we use the notation $[p]\in <!--\; \in \; -->E$ if the projectivization of this point lies on the curve. With this notation, you have $[\mathrm{\lambda <!--\; \lambda \; -->}p]=[p]$ for any $\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->k^{\times <!--\; \times \; -->}$. Let $p\in <!--\; \in \; -->V$ be such that $\mathrm{\theta <!--\; \theta \; -->}=[p]$ is the distinguished point of $E$. Note that:
$\begin{array}{rl}\mathrm{\phi <!--\; \phi \; -->}([p])& =[x([p]):y([p]):1]\\ & =[x([\mathrm{\lambda <!--\; \lambda \; -->}p]):y([\mathrm{\lambda <!--\; \lambda \; -->}p]):1]\\ & =[{\mathrm{\lambda <!--\; \lambda \; -->}}^{-<!--\; -\; -->2}\cdot <!--\; \cdot \; -->x([p]):{\mathrm{\lambda <!--\; \lambda \; -->}}^{-<!--\; -\; -->3}\cdot <!--\; \cdot \; -->y([p]):1]\\ & =[\mathrm{\lambda <!--\; \lambda \; -->}x([p]):y([p]):{\mathrm{\lambda <!--\; \lambda \; -->}}^3]\end{array}$
Because $x$ and $y$ have the aforementioned order at $[p]$. Now if you let $\mathrm{\lambda <!--\; \lambda \; -->}$ approach zero in this expression on the right (which is constant!), the point approaches $[0:1:0]$. This can be turned into a formal proof if your field is actually continuous, but I will not spend a lot of time on it because we don't have or need this assumption here.
Rigorous Approach
To prove this properly, we need to understand the morphism globally, and provide a different rational representation around $\mathrm{\theta <!--\; \theta \; -->}$. The morphism φ corresponds to a morphism of (function) fields ${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}:k(X,Y,Z)\to <!--\; \to \; -->k(E)$. The distinguished point $\mathrm{\theta <!--\; \theta \; -->}$ is a maximal ideal in $k[E]$ and we can write
1. ${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}(X)=x=\frac{a}{b}$
2. ${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}(Y)=y=\frac{c}{d}$
While $\mathrm{\zeta <!--\; \zeta \; -->}:={\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}(Z)$ is some element that is not in $\mathrm{\theta <!--\; \theta \; -->}$, we have $a,b,c,d\in <!--\; \in \; -->\mathrm{\theta <!--\; \theta \; -->}$ and as per the assumption:
1. $v_{\mathrm{\theta <!--\; \theta \; -->}}(a)-<!--\; -\; -->v_{\mathrm{\theta <!--\; \theta \; -->}}(b)=-<!--\; -\; -->2$
2. $v_{\mathrm{\theta <!--\; \theta \; -->}}(c)-<!--\; -\; -->v_{\mathrm{\theta <!--\; \theta \; -->}}(d)=-<!--\; -\; -->3$
Now let's define an a-priori different rational map $f:k(X,Y,Z)\to <!--\; \to \; -->k(E)$ as the composition of ${\mathrm{\phi <!--\; \phi \; -->}}^{\ast <!--\; \ast \; -->}$ with the multiplication by ${\mathrm{\lambda <!--\; \lambda \; -->}}^3$ where $\mathrm{\lambda <!--\; \lambda \; -->}$ is a function with $v_{\mathrm{\theta <!--\; \theta \; -->}}(\mathrm{\lambda <!--\; \lambda \; -->})=1$. In other words:
1. $f(X):=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{3}a}{b}$
2. $f(Y):=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{3}c}{d}$
3. $f(Z):={\mathrm{\lambda <!--\; \lambda \; -->}}^3\mathrm{\zeta <!--\; \zeta \; -->}$
Hence:
1. $v_{\mathrm{\theta <!--\; \theta \; -->}}(f(X))=3+v_{\mathrm{\theta <!--\; \theta \; -->}}(t)-<!--\; -\; -->v_{\mathrm{\theta <!--\; \theta \; -->}}(u)=1$ (first coordinate vanishes)
2. $v_{\mathrm{\theta <!--\; \theta \; -->}}(f(Y))=3+v_{\mathrm{\theta <!--\; \theta \; -->}}(v)-<!--\; -\; -->v_{\mathrm{\theta <!--\; \theta \; -->}}(w)=0$ (second coordinate doesn't)
3. $v_{\mathrm{\theta <!--\; \theta \; -->}}(f(Z))=3$ (third coordinate vanishes)
With $\mathrm{\psi <!--\; \psi \; -->}:=f^{\ast <!--\; \ast \; -->}$, this implies $\mathrm{\psi <!--\; \psi \; -->}(\mathrm{\theta <!--\; \theta \; -->})=[0:1:0]$. Now we are left to verify that the rational functions $\mathrm{\psi <!--\; \psi \; -->}$ and $\mathrm{\phi <!--\; \phi \; -->}$ are the same, which would prove that $\mathrm{\psi <!--\; \psi \; -->}$ is an extension of $\mathrm{\phi <!--\; \phi \; -->}$ to $\mathrm{\theta <!--\; \theta \; -->}$ which proves what we desire.
This is fairly straightforward: We only need to check that they agree on an open subset. For this subset, simply choose one where all of the functions $\mathrm{\zeta <!--\; \zeta \; -->},\mathrm{\lambda <!--\; \lambda \; -->},a,b,c,d$ are nonzero and you will get that the projective coordinates of $\mathrm{\psi <!--\; \psi \; -->}(P)$ and $\mathrm{\phi <!--\; \phi \; -->}(P)$ differ by the nonzero scalar factor ${\mathrm{\lambda <!--\; \lambda \; -->}}^3(P)$, and so they are identical.