In page number 3 of J.S. Golan's book Semirings and their applications, there is a result which says that
if $R$ is a hemiring and $S$ is a subhemiring of $R$ which is a semiring having multiplicative identity $e$ then the set $R\times S$, on which the operations of addition and multiplication are defined by $$(r,s)+(r_1,s_1)=(r+r_1,s+s_1)$$ and $$(r,s)(r_1,s_1)=(rr_1+rs_1+sr_1,ss_1)$$ is a semiring with unity $(0,e)$.
But I have a question: after $(r,s)(0,e)=(re,s)$, how can we say that $re=r?$
It needs to be posited that $er=re=r$ for all $r\in R$ for this to work.
If this requirement that $er=re=r$ isn't met, it's easy to find counterexamples to the statement even for rings.
Take $R=F\times F$ for a field $F$ and $S=\{0\}\times F$.
$S$ is a ring with identity $e=(0,1)$, and clearly it is not true that $er=re=r$ for all $r$. It doesn't work for $(1,1)$, for example.
This construction is usually called the Dorroh extension when $S=\Bbb Z$. In that case, $1r=r1=r$ for all $r$ since integer multiplication is interpreted as repeated addition in the ring's abelian operation.