Added: This answer the updated question.
Your updated conjecture is also false. Consider:
$$
A = \left(
\begin{array}{cc}
10 & -5 \\
9 & 3 \\
\end{array}
\right) \qquad\qquad\qquad
B = \left(
\begin{array}{cc}
1 & -6 \\
8 & 8 \\
\end{array}
\right)
$$
Then
$$
\frac{1}{2}(A.B + B.A) - 2 ( (A.B)^{-1} + (B.A)^{-1} )^{-1} = \frac{1}{7474} \left(
\begin{array}{cc}
-98938 & -164451 \\
247345 & -61502 \\
\end{array}
\right)
$$
The matrix above is negative-definite.
Added: This portion answers the first variant of the problem:
No, it is not true. Counterexample:
$$
A = \left(
\begin{array}{ccc}
3 & 3 & -4 \\
1 & -2 & 0 \\
-1 & -2 & -3 \\
\end{array}
\right)
$$
Then
$$
\frac{1}{2}( A + A^t) - 2 ( A^{-1} + (A^{-1})^t)^{-1} = \left(
\begin{array}{ccc}
\frac{127}{198} & \frac{4}{33} & -\frac{163}{198} \\
\frac{4}{33} & -\frac{2}{11} & \frac{5}{33} \\
-\frac{163}{198} & \frac{5}{33} & \frac{59}{99} \\
\end{array}
\right)
$$
It eigenvalues are, approximately, $1.44$, $-0.387$ and zero.
Added: In order to create such a counterexample I have used Mathematica:
f[a_] := 1/2 (a + Transpose[a]) -
2 Inverse[# + Transpose[#] &[Inverse[a]]]
Now this generates random integer-valued matrices until a non-degenerate one is generated with the combination in question having eigenvalues of opposite signs:
While[True,
While[Det[a = RandomInteger[{-4, 4}, {2, 2}]] == 0, Null];
If[ Intersection[Sign[Eigenvalues[f[a]]], {-1, 1}] == {-1, 1},
Break[]]
];
Here is a screenshot:
