Maybe it's time for a (reasonably complete) answer.
Consider the open subvariety $U_{YZ}:=\Bbb{P}^2 \setminus V(YZ)$ (i.e. $\Bbb{P}^2$ with both the $Y$-axis and the $Z$-axis removed). The rational map $f: \Bbb{P}^2 \dashrightarrow \Bbb{P}^2$ can be represented on $U_{YZ}$ by the morphism
\begin{split}
U_{YZ} &\rightarrow U_Z:=\Bbb{P}^2 \setminus V(Z)\\
(x:y:z) &\mapsto (x/z:x/y:1)
\end{split}
which shows that $f$ is defined on all of $U_{YZ}$. Similarly, you find that $f$ is defined on all of $\Bbb{P}^2 \setminus \{(1:0:0),(0:1:0),(0:0:1)\}$, and the only thing left to check for 1. and 2. is that $f$ cannot be extended to all of $\Bbb{P}^2$. One possible explanation is the following.
Consider the morphism
\begin{split}
f_{cone}: \Bbb{A}^3 &\rightarrow \Bbb{A}^3\\
(x,y,z) &\mapsto (xy,xz,yz)
\end{split}
which maps every point of the form $(\lambda,0,0),(0,\mu,0)$ or $(0,0,\nu)$ to the point $(0,0,0)$.
Like every morphism of varieties, $f_{cone}$ is already uniquely determined by its restriction to any non-empty open subset, e.g. by its restriction to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\} \subset \Bbb{A}^3$.
But $f_{cone}$ restricted to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\}$ induces the morphism representing the rational map $f$, as described above. Hence every extension of $f$ to a morphism defined on all of $\Bbb{P}^2$ would have to map the points $(1:0:0),(0:1:0)$ and $(0:0:1)$ to "$(0:0:0)$", which is impossible, and consequently, the points $(1:0:0),(0:1:0),(0:0:1)$ cannot be in the domain of $f$.
For 3., you simply use the "trick" mentioned by Asal. Let $U_{XYZ}:=\Bbb{P}^2 \setminus V(XYZ)$ (i.e. $\Bbb{P}^2$ without the coordinate axes). On $U_{XYZ}$, we can represent $f$ as
$$
(x:y:z) \mapsto (xy:xz:yz)=1/(xyz)(xy:xz:yz)=(1/z:1/y:1/x)
$$
and from this, you can read off the inverse of $f_{\vert U_{XYZ}}$ directly.
First, we note that $\Bbb P^2$ and $X$ are birational: the open sets $\Bbb P^2\setminus P$ and $X\setminus E$ are isomorphic (the blowup map actually restricts to the identity here). This means that the function fields are isomorphic, and a function $f$ which has $div(f)=E$ on $X$ would necessarily have $div(f)=P$ on $\Bbb P^2$, which is obviously impossible. So $E\neq 0$ in the Picard group.
Tabes Bridges' comment is also a possible approach to the problem, as the self-intersection number can be computed only by knowing the degree of the normal bundle (without assuming anything about the class of $E$). As the exceptional divisor $E$ of a blowup in a point of a smooth variety of dimension $n$ has normal bundle $\mathcal{O}_E(E) \cong \mathcal{O}_{\Bbb P^{n-1}}(-1)$ (this is standard when defining the blowup and does not require knowledge of the Picard group of the blowup), we see that the self-intersection number of the exceptional divisor is $(-1)^{n-1}$. So if one can demonstrate that the intersection product is a well-defined thing on the Picard group, this would suffice to show that $E\neq 0$.
Best Answer
Since the coordinates on $\mathbb P^1$ are homogeneous the map you've written: $$ (x,t) \overset{p|_U}{\longmapsto} (x:tx) $$ is the same as $$ (x,t) \overset{p|_U}{\longmapsto} (1:t). $$ Meaning that they coincide on the dense open set $$ U\cap V = V(z,s)^c\cap V(z,t)^c=\{V(z,s)\cup V(z,t)\}^c=V(z,s,t)^c=\{z\neq0,s\neq0,t\neq0\}, $$ so to extend the definition everywhere we just use the second definition.
Another reason the map has to be defined everywhere is that it is the restriction of the projection $\mathbb P^2\times \mathbb P^1\to \mathbb P^1$.