Write $\mathbb{CP}^1$ as two copies of $\mathbb{C}$, with coordinates $z_1$ and $z_2$ respectively, glued along the map
$z_1 \mapsto z_2=\frac{1}{z_1}$ on $z_1\neq 0$.

Write the line bundle $\mathcal{O}(-1) \rightarrow \mathbb{CP}^1$ as two copies of $\mathbb{C} \times \mathbb{C}$, with coordinates $(z_1, v_1)$ and $(z_2,v_2)$ respectively, glued along the map $(z_1,v_1) \mapsto (z_2, v_2) = (\frac{1}{z_1}, z_1 v_1)$ on $z_1\neq 0$. Denote the zero-section by $Z$.

By definition of a blow-up, there is a holomorphic isomorphism from a small neighborhood of E to a neighborhood of $Z$, and this isomorphism sends $E$ to $Z$.

Therefore, it is enough to compute the self-intersection of $Z$. This is a topological notion, so a natural thing to do is to find a cycle $\gamma$ homologous to $Z$ and intersecting $Z$ transversely. (You can't ask $\gamma$ to be a divisor: $Z$ is the only compact divisor in $\mathcal{O}(-1)$.)

Construct such a $\gamma$ as *continuous* section of $\mathcal{O}(-1)$:
on $\vert z_1 \vert \leq 1$, take $z_1 \mapsto (z_1,v_1=1)$; on $\vert z_2 \vert \leq 1$, take $z_2 \mapsto (z_2,v_2= \overline{z_2})$. On the overlap $\vert z_1 \vert=1= \vert z_2 \vert$, we have
$v_2= \overline{z_2} = \frac{1}{z_2} = z_1 = z_1 v_1$, as needed.

A homotopy from $Z$ to $\gamma$ is given by
$z_1 \mapsto (z_1,t)$ and $z_2 \mapsto (z_2, t\ \overline{z_2})$, for $t\in [0,1]$.
In particular (the image of) $\gamma$ is homologous to $Z$.

The only intersection point of $\gamma$ and $Z$ is at $(z_2,v_2)=(0,0)$. There, the orientation of $Z$ given by its complex structure is represented by the vectors $(1,0)$ and $(i,0)$. Pushing this orientation with the above homotopy gives the orientation for $\gamma$ represented by $(1,1)$ and $(i,-i)$.

The $\mathbb{R}$-basis of $\mathbb{C} \times \mathbb{C}$ given by $(1,0)$, $(i,0)$, $(1,1)$ and $(i,-i)$ has same orientation as
$(1,0)$, $(i,0)$, $(0,1)$ and $(0,-i)$, which is negative. Conclusion: $Z.\gamma = -1$, so
$Z.Z=-1$, so $E.E=-1$.