= Solution
Since $\sigma$ is a <group homomorphism>, $\sigma(0)=I$, $\sigma(x+y)=\sigma(x)\sigma(y)$, and $\sigma(-x)=\sigma(x)^{-1}$. The product is closed on $\mathbb R\times\mathbb R^2$. Its two associative bracketings give
$$
\begin{aligned}
[(x,v)*(y,w)]*(z,u)
&=(x+y+z,v+\sigma(x)w+\sigma(x+y)u),\\
(x,v)*[(y,w)*(z,u)]
&=(x+y+z,v+\sigma(x)w+\sigma(x)\sigma(y)u),
\end{aligned}
$$
and coincide by the homomorphism property. The <identity element> is $(0,0)$, and the two-sided inverse is
$$
\boxed{(x,v)^{-1}=(-x,-\sigma(-x)v)}.
$$
For example, multiplying on the right gives $v-\sigma(x)\sigma(-x)v=0$, and multiplying on the left gives $-\sigma(-x)v+\sigma(-x)v=0$. Thus this law defines a <group>, specifically a <semidirect product>.
If it is an <abelian group>, comparing $(x,0)*(0,w)$ and $(0,w)*(x,0)$ gives $\sigma(x)w=w$ for every $w$. Hence $\sigma(x)=I$ for every $x$. Conversely this trivial action makes the product ordinary vector addition. Therefore \b[the unique action giving a commutative group is]
$$
\boxed{\sigma(x)=I\quad\text{for all }x}.
$$
For the final condition, compare $(0,ae_1)*(x,w)$ and $(x,w)*(0,ae_1)$. They are $(x,ae_1+w)$ and $(x,w+\sigma(x)ae_1)$ respectively. They coincide for every $a,x,w$ exactly when $\sigma(x)e_1=e_1$ for every $x$. The first column of $\sigma(x)$ is then $(1,0)^{\mathsf T}$; since it lies in the <special linear group>, its determinant is one and its other diagonal entry is also one. Thus
$$
\sigma(x)=\begin{pmatrix}1&r(x)\\0&1\end{pmatrix}.
$$
Matrix multiplication gives $r(x+y)=r(x)+r(y)$. Conversely any such <additive function> gives a homomorphism into $SL(2,\mathbb R)$ fixing $e_1$ and therefore makes these translations central. This proves the full characterization of the <central translations in a linear semidirect product>:
$$
\boxed{r:(\mathbb R,+)\longrightarrow(\mathbb R,+)\text{ may be any homomorphism}}.
$$
No continuity or measurability is assumed, so it would be unjustified to restrict the answer to $r(x)=cx$.
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