= Solution
For $a,b\in I$, the <square-zero ideal> condition gives
$$
(1+a)(1+b)=1+a+b+ab=1+a+b,
\qquad (1+a)^{-1}=1-a.
$$
Thus the <square-zero unit subgroup> $1+I$ is abelian, and
$$
I_{\mathrm{add}}\longrightarrow1+I,
\qquad a\longmapsto1+a
$$
is a <group isomorphism> from the <additive group> of $I$.
Use the specified <ring isomorphism> $R/I\cong\mathbb ZG$. For $x\in\mathbb ZG$, choose a lift $r\in R$ and define $x\cdot a=ra$ for $a\in I$. Two lifts differ by an element of $I$, whose product with $a$ vanishes, so this is well defined. Right multiplication is handled identically. The two actions commute by <associativity>, making $I$ a <bimodule>. If $u\in R$ lifts $g\in G$, then $u$ is a <unit>: a lift $v$ of $g^{-1}$ makes both $uv$ and $vu$ elements of $1+I$, hence units, and a ring element with both a left and a right inverse is invertible. Conjugation therefore defines
$$
g\cdot a=uau^{-1}.
$$
Changing $u$ by an element of $I$ does not change this expression because $I^2=0$. Moreover,
$$
u(1+a)u^{-1}=1+uau^{-1},
$$
so $a\mapsto1+a$ is an isomorphism of $\mathbb ZG$-modules for these <conjugation actions>.
Let $R^\times\to(R/I)^\times$ be reduction on <unit groups>, and define $U$ as the inverse image of the distinguished subgroup $G\subseteq(\mathbb ZG)^\times$. Every $g\in G$ has a unit lift by the preceding argument, and the kernel consists exactly of the units congruent to $1$, namely $1+I$. Multiplication in $R$ therefore gives the <group extension>
$$
1\longrightarrow1+I\longrightarrow U\longrightarrow G\longrightarrow1.
$$
Choose a set-theoretic section $s:G\to U$ with $s(1)=1$. Its <extension cocycle>
$$
c(g,h)=s(g)s(h)s(gh)^{-1}\in1+I
$$
satisfies the <two-cocycle> identity by <associativity>. A different section changes $c$ by a <group coboundary>, so <second group cohomology classifies group extensions> gives a well-defined class
$$
x=[c]\in H^2(G,1+I).
$$
The same construction for $(R_1,I_1)$ gives $U_1$, an extension cocycle $c_1$, and $x_1=[c_1]\in H^2(G,1+I_1)$.
The answer to the final question is no. An abstract <ring isomorphism> $R_1\cong R$ need not carry the distinguished ideal $I_1$ to $I$, need not induce the identity under the two chosen identifications of the <quotient rings> with $\mathbb ZG$, and need not induce the prescribed $\mathbb ZG$-module isomorphism $\theta$. Hence it need not give an isomorphism of the two displayed <group extensions>, so it imposes no equality $\phi(x_1)=x$. That equality does hold if the ring isomorphism has all three compatibility properties, because it then carries one extension cocycle to the other up to a <group coboundary>.
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