= Solution
Define
$$
\alpha:E\longrightarrow\operatorname{Cl}^0(E),
\qquad
\alpha(P)=[P-Q].
$$
If $D$ is a <divisor class> of degree zero, <Riemann-Roch theorem>[Riemann--Roch] gives $\ell(D+Q)=1$, because $K_E\sim0$ and the divisor $-D-Q$ has negative degree. A nonzero section therefore yields
$$
D+Q\sim P
$$
for some point $P$, proving that $\alpha$ is surjective. If $\alpha(P)=\alpha(P')$, then $P-P'$ is a <principal divisor>. For $P\ne P'$ its defining function would have one simple pole and give a degree-one map $E\to\mathbb P^1$, impossible for a genus-one curve. Thus $P=P'$. The map and its inverse are regular, so this proves the <Abel-Jacobi map of a genus-one curve> is an isomorphism.
Transport the addition of the <divisor class group> through $\alpha$: define $P\boxplus R$ by
$$
[P\boxplus R-Q]=[P-Q]+[R-Q].
$$
Then $Q$ is the identity, and this is the <chord-and-tangent group law> on the plane cubic.
For the displayed equation, inversion is $(x,y)\mapsto(x,-y)$. Thus $P\boxplus P=Q$ exactly when $P=-P$. Besides the point at infinity $Q$, this requires $y=0$. Therefore the complete set of two-torsion points is
$$
\boxed{Q,\quad(\lambda_1,0),\quad(\lambda_2,0),\quad(\lambda_3,0).}
$$
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