= Solution
Use <left Grassmann derivatives> and place the odd constant transformation parameters on the left. A <scalar superfield transformation> can be written
$$
\delta S=(\epsilon^\alpha Q_\alpha+\bar\epsilon^{\dot\alpha}\bar Q_{\dot\alpha})S,
\qquad
Q_\alpha=\partial_\alpha-i\sigma^\mu_{\alpha\dot\beta}\bar\theta^{\dot\beta}\partial_\mu,
\qquad
\bar Q_{\dot\alpha}=\bar\partial_{\dot\alpha}-i\theta^\beta\sigma^\mu_{\beta\dot\alpha}\partial_\mu.
$$
Equivalently, pull back the scalar function under the infinitesimal <superspace> translation
$$
\delta\theta=\epsilon,\qquad\delta\bar\theta=\bar\epsilon,
\qquad\delta x^\mu=i\theta\sigma^\mu\bar\epsilon-i\epsilon\sigma^\mu\bar\theta.
$$
The term involving $\bar\epsilon$ has the displayed sign because interchanging it with $\theta$ reverses the sign in the <Grassmann algebra>. Overall factors of $i$ in the definition of the <supercharges> can be changed together without changing this coordinate transformation.
For two even scalar <superfields> $S,T$, the odd generators obey the <graded Leibniz rule>, while the full parameter-weighted variation $\delta$ is even. Thus
$$
\delta(ST)=(\delta S)T+S(\delta T).
$$
This is precisely the transformation of a <product of scalar superfields>. The finite version is even clearer: if $S'(z)=S(z')$ and $T'(z)=T(z')$ for the same transformed <superspace> point, then $(ST)'(z)=S(z')T(z')=(ST)(z')$. No new representation law is required.
The expression $S(x,\theta,\bar\theta)=\phi(x)$ needs a distinction between a configuration and a closed field content. As a function on <superspace>, it is a valid special configuration of an unconstrained <superfield>. But its variation is
$$
\delta S=(i\theta\sigma^\mu\bar\epsilon-i\epsilon\sigma^\mu\bar\theta)\partial_\mu\phi.
$$
For nonconstant $\phi$, this generates nonzero components depending on <Grassmann variables>. Therefore \b[the pure-scalar truncation is not a supersymmetric multiplet]: an isolated spacetime <scalar field> with all its partners permanently set to zero is not closed under <supersymmetry>. This is the <pure-scalar truncation of a superfield>. A constant gauge-singlet $\phi$ is the trivial invariant exception. Calling $\phi(x)$ a special <superfield> configuration is consistent; claiming that its restricted field content remains a <superfield> representation is not.
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