(1) This is the Principle of Inclusion-Exclusion from basic combinatorics.
(2) Set $ A := I(G) $. Let $ x \in I(G) $. As left-multiplication by a group element is an injective operation on the group, we see that $ B := xA $ has the same cardinality as $ A $. Hence,
$$
|A| = |B| \geq \frac{3}{4} |G|.
$$
Using (1), we thus obtain
\begin{align}
|I(G) \cap x[I(G)]| &= |A \cap B| \\
&\geq |A| + |B| - |G| \\
&\geq \frac{3}{4} |G| + \frac{3}{4} |G| - |G| \\
&= \frac{1}{2} |G|.
\end{align}
(3) Let $ x \in I(G) $, and note that $ 1_{G} \in {C_{G}}(x) $. Suppose that $ g \in I(G) $ satisfies $ xg \in I(G) $. Then
\begin{align}
xg &= (xg)^{-1} \quad (\text{As $ (xg)^{2} = 1_{G} $.}) \\
&= g^{-1} x^{-1} \\
&= gx, \quad (\text{As $ g^{2} = 1_{G} = x^{2} $.})
\end{align}
which yields $ g \in {C_{G}}(x) $. Hence,
$$
\{ 1_{G} \} \cup \{ g \in I(G) ~|~ gx \in I(G) \} \subseteq {C_{G}}(x).
$$
(4) Let $ x \in I(G) $. Observe that
\begin{align}
|{C_{G}}(x)|
&\geq |\{ 1_{G} \} \cup \{ g \in I(G) ~|~ gx \in I(G) \}| \quad (\text{By (3).}) \\
&= |\{ 1_{G} \} \cup (I(G) \cap x^{-1} [I(G)])| \\
&= |\{ 1_{G} \}| + |I(G) \cap x^{-1} [I(G)]| \quad (\text{As $ 1_{G} \notin I(G) $.}) \\
&\geq 1 + \frac{1}{2} |G| \quad (\text{Applying (2) to $ x^{-1} $, which is an element of $ I(G) $.}) \\
&> \frac{1}{2} |G|.
\end{align}
Hence, $ [G:{C_{G}}(x)] < 2 $, which yields $ {C_{G}}(x) = G $, or equivalently, $ x \in Z(G) $.
As $ x \in I(G) $ is arbitrary, we have $ I(G) \subseteq Z(G) $. Then as $ |I(G)| \geq \dfrac{3}{4} |G| $, it follows that
$$
[G:Z(G)] \leq \frac{4}{3} < 2.
$$
Therefore, $ Z(G) = G $.
Conclusion: $ G $ is an abelian group.
