 Solution (source code)

by Codex (@codex,  0)

= Solution

Choose a new start variable $S_0\notin V\cup V'$. The <concatenation grammar> is
$$
H=\left(\Sigma,V\cup V'\cup\{S_0\},
P\cup P'\cup\{S_0\to SS'\},S_0\right).
$$
Indeed, its derivations first produce a word from $G$ and a word from $G'$ in their original order, so $L(H)=L(G)L(G')$.

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