It's a well-known fact that if $A$ an Abelian group and $G$ is a group, then all group extensions of $G$ by $A$ is isomorphic with the group ($A\times G,\,\bullet)$, where the group operation $\bullet$ is
$$(a_1,g_1)\bullet(a_2,g_2) = (a_1+\varphi_{g_1}(a_2)+f(g_1,g_2),\,g_1g_2)\tag{1}$$
where
- $\varphi:(A\times G)\to A: (a,g)\mapsto \varphi_g(a)$ is a group action of $G$ on $A$
- $f: G\times G\to A$ is a cocycle, i.e. satisfies $f(g_1,\,g_2g_3)+\varphi_{g_1}(f(g_2,\,g_3)) = f(g_1g_2,\,g_3)+f(g_1,\,g_2)$.
I'd like to compute the inverse element of $(a,g)$. For the sake of simplicity, let's take a normalized $f$, that is, for the identity element $e$ of $G$ suppose $f(e,e)=0$. In this case, $$f(g,e)=f(e,g)=f(e,e)=0\tag{2}$$ for all $g\in G$ (where $0$ is the identity element of $A$), and the identity element of $(A\times G,\bullet)$ is $(0,e)$. So, if $(a,g)^{-1}=(a_1,g_1)$ then $$ (a_1,g_1)(a,g) = (0,g)\tag{3}$$ and $$ (a,g)(a_1,g_1) = (0,g)\tag{4}$$ From (1), (2) and (3) $$g_1=g^{-1}$$ and $$a_1=\varphi_{g}(a)-f(g^{-1},g)\tag{5}$$ while from (1), (2) and (4) $$ a_1=\varphi_g(a)+\varphi_g(f(g,g^{-1}))\tag{6}$$
From the equality of the RHS of (5) and (6)
$$-f(g^{-1},g)=\varphi_g(f(g,g^{-1}))\tag{7}$$
Is it sure that this holds for every normalized 2-cocycle? I couldn't derive this from the cocycle condition
$$ f(g_1,g_2g_3)+\varphi_{g_1}(f(g_2,g_3)) = f(g_1g_2,g_3) + f(g_1,g_2)\tag{8}$$ Perhaps I missed something?
(I think in (3) and (4), the "$(0,g)$" should be "$(0,e)$".)
I am not sure how you are getting (5) and (6). Using (1) and (3), I get \begin{align} a_{1}+\varphi_{g^{-1}}(a) + f(g^{-1},g) = 0 \tag{5$'$} \end{align} and using (1) and (4), I get \begin{align} a+\varphi_{g}(a_{1}) + f(g,g^{-1}) = 0 \tag{6$'$} \end{align} and applying $\varphi_{g^{-1}}$ to (6') gives \begin{align} \varphi_{g^{-1}}(a)+a_{1} + \varphi_{g^{-1}}(f(g,g^{-1})) = 0 \tag{6$''$} \end{align} and comparing (5') and (6'') gives \begin{align} f(g^{-1},g) = \varphi_{g^{-1}}(f(g,g^{-1})) \tag{7$'$} \end{align} which you can get by setting $(g_{1},g_{2},g_{3}) = (g^{-1},g,g^{-1})$ in (8) and using (2).