Let $R \neq 0$ be a commutative ring, $G$ be a finite group, $\#G > 2$.
1) $H$ subgroup of G $\Rightarrow$ monoid ring R[H] is a subring of monoid ring R[G]
2) Let $x := \sum_{g\in G}g, \ X := \left\{ r x + r'e_G \mid r,r' \in R \right\} \Rightarrow$ $X$ subring of $R[G]$ and there is no subgroup H of G, so that $X = R[H]$.
1)
As far as I understand, I can follow associativity for both operations from $H \leq G$, and do only need to show that the operations are closed, am I right?
Let $x,y \in R[H] \Leftrightarrow x = \sum_{h\in H}a_h h, y = \sum_{h\in H}b_h h$.
$$x + y = \sum_{h \in H} (\underbrace{a_h + b_h}_{\in H}) h \in R[H]$$
$$x y = \sum_{h\in H}\left(\sum_{h_1,h_2 \in H, h_1h_2=h} \underbrace{a_{h_1} b_{h_2}}_{\in H} \right)h \in R[H]$$
Is this all that need to be shown here? Maybe it is left to show that the sums are finite, but that should directly follow from the fact, that the sums of x and y are finite, am I right?
2)
How do I show that there is no such subgroup? I played around with the definitions but didn't find an approach, can you please help me to find a way?
Regarding 1)
Usually the subring test involves closure of multiplication and subtraction (but this isn't too different from addition here). Also $a_h+b_h$ is an element of the ring $R$, not of the group $H$, so your notation is a bit confusing. The point is that a general element of $R[H]$ is a formal sum of elements of $H$ with arbitrary coefficients in $R$ and after multiplication or subtraction of two elements of $R[H]$ the result is again a formal sum which only involves elements of $H$.
Regarding 2)
You can think of the set $X$ as vectors $(a,b,b,\dots)$ with $a,b \in R$. All group elements have the same coefficient, except the identity element, which can be anything. Now for any subgroup $H$ $R[H]$ contains the element $(0,1,0,\dots)$. Can this element be in $X$ if $\#G>2$?