I'll take "inversable" as meaning "invertible". Let $a={\rm tr}(A),\ d=\det(A)$, then the hypothesis are $a,d\neq0,$ and the characteristic polynomial of $A$ is $t^4-at^3+bt^2-at+d$ for some $b\in{\mathbb R}.$ Suppose there exists $B\in\mathcal{M}_4(\mathbb{R})$ such that $B\neq0$ and $AB=-BA.$ Then $B$ anti-commutes with all the odd powers of $A$, while it commutes with all the even powers of $A.$ Note that, on a filed of characteristic different from 2, if a matrix $M$ commutes and anti-commutes with a matrix $N$, then $MN=NM=0$ ($MN=NM=-MN$.) We have $A^4+bA^2+dI=a(A^3+A)$ so $B$ commutes with $A^4+bA^2+dI$ and anti-commutes with $a(A^3+A)$, thus $aA(A^2+I)B=0$, which implies $(A^2+I)B=0$, since $a\neq0$ and $A$ is invertible. We obtain that the image of $B$ (which is not $\{0\}$) is included in the kernel of $A^2+I$, which is therefore singular. Working in the other direction, we don't need all the hypothesis, just that $a\neq 0$. Observe that the problem is invariant under similarity, so we may suppose $A$ in (real) Jordan normal form. Suppose $A^2+I$ is singular. Then $A^2$ has the eigenvalue -1, which implies that the (complex) spectrum of $A$ contains $i$ and $-i$. Since $a\neq0$, the spectrum of $A$ can't contain only $i$ and $-i$. This in turn implies that the Jordan normal form of $A$ is of type $\begin{pmatrix}C&0\\ 0&D\end{pmatrix}$ where $C=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}$ and $D\in\mathcal{M}_2(\mathbb{R})$. Then take $B=\begin{pmatrix}E&0\\ 0&0\end{pmatrix}$ with $E=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$.