Edited. (Thanks to Ravi for spotting the original error!)
As Ravi pointed out, the embedding of the abstract elliptic curve in $\mathbb P^2$ can be described as follows: one picks a point $P_0$ on the curve, and embedds using the complete linear system $|3P_0|$. Once embedded in $\mathbb P^2$, the divisor $3P_0$ is then equivalent to the class of the hyperplane $H$ associated with this embedding in $\mathbb P^2$.
In your case, I believe that $P_0 = [0 : 1 : 0]$ is the point that Hartshorne has in mind for this purpose. The fact that $3P_0 \sim H$ is clear from the fact that the hyperplane $z = 0$ has triple intersection with the elliptic curve at $P_0$.
So having identified $3P_0 \sim H$, one can use the following statement to characterise the group law:
$$ P + Q + R = 0 \iff P + Q + R \sim H $$
For the purposes of your exercise from Hartshorne, I would compute $P + Q$ in two stages. First, I would determine $S := - (P + Q)$ by finding the hyperplane that intersects the elliptic curve at $P$ and $Q$ and taking $S$ to be the third point of intersection. Second, I would determine $R := - S = - ( P_0 + S)$ by finding the third intersection point for the hyperplane that intersects the elliptic curve at $S$ and $P_0$ and taking $R$ to be the third point of intersection.
Finally, note that if we are merely presented with the embedding of the elliptic curve in $\mathbb P^2$, then $P_0$ is not uniquely determined. Indeed if $3P_0 \sim H$, then we also have $3P'_0 \sim H$ for any $P'_0$ that is a $3$-torsion point with respect to the group law defined with origin $P_0$. Thanks again to Ravi for pointing this out!