Let me prove a lemma first.
Lemma : Let $M\in GL_n(\mathbb{C}), \exists P\in \mathbb{C}[X]$ such that $M=\exp(P(M))$.
With Dunford decomposition $M=D+N$, where $D,N$ are polynomials in $M$ and commute. Thus it exist $P$ a complex polynom such that : $$D=Pdiag(\lambda_1,\dots,\lambda_n)P^{-1}$$
The complex exponential is surjective thus $\forall i,\exists \nu_i$ such that $e^{\nu_i}=\lambda_i$. Let $Q$ the Lagrange polynomial such that $Q(\lambda_i)=\nu_i$.
$$Q(D)=PQ(diag(\lambda_1,\dots,\lambda_n))P^{-1}=Pdiag(\nu_1,\dots,\nu_n)P^{-1}$$
$$\exp(Q(D))=Pdiag(e^{\nu_1},\dots,e^{\nu_n})P^{-1}=D$$.
$Q(D)$ is a polynomial in $M$, so the lemma is true if the matrix is diagonalisable.
Also $M$ is invertible and $D$ has the same eigen values so $D$ is invertible. So $(D^{-1}N)^k=D^{-k}N^k$, $N$ is nilpotent so $D^{-1}N$ is also nilpotent. Thus :
$$I_n+D^{-1}N=\exp(\sum_{k=1}^n \frac{(-1)^{k-1}}{k}(D^{-1}N)^k)$$
Also we have the characteristic polynomial $$\chi_D=(-1)^nX^n+\sum_{k=0}^{n-1}a_kX^k$$
$a_0\neq0$, because $D$ is invertible, so with Cayley-Hamilton theorem :
$$D^{-1}=-\frac{1}{a_0}((-1)^nD^{n-1}+\sum_{k=1}^{n-1}a_kD^{k-1})$$
$D$ and $N$ are polynomials in $M$ so $\exists B\in \mathbb{C}[X]$ such that $I_n+D^{-1}N=\exp(B(M))$.
Finally :
$$M=D(I_n+D^{-1}N)=\exp(A(M))\exp(B(M))=\exp((A+B)(M))$$
because $A(M)$ and $B(M)$ commute. The lemma is proved.
Now we will prove that $\exp(M_n(\mathbb R))=\{A\in GL_n(\mathbb{R}) | \exists B \in M_n(\mathbb R),A=B^2\}$.
Let $A \in \exp (Mn(\mathbb{R}))$, then $\exists B \in Mn(\mathbb{R})$ such that $A=\exp(B)$, so $A = (\exp(\frac 1 2 B))^2$ is the square of an invertible matrix so it is invertible.
Let show the other inclusion. If $A\in GLn(\mathbb R))$ such that $\exists B \in M_n(\mathbb R)$ and $A=B^2$, so we have $\det(B)^2=\det(A)\neq 0$ so $B$ is invertible. By the lemma $\exists P \in \mathbb C [X]$ such that $B=\exp(P(B))$. Since $B$ is real we have $A=B\overline B=\exp(P(B))\overline{\exp(P(B))}=\exp(P(B))\exp(\overline P(B))=\exp((P + \overline P)(B))$ because $P(B)$ and $\overline P (B)$ commute. And since $P+\overline P$ is a real polynomial we have $A\in \exp(Mn(\mathbb R))$.
We have proved the equality of the two sets.
Finaly we can prove that $\{A \in GL_n(\mathbb{R}) | \exists B \in M_n(\mathbb R),A=B^2\} \neq GLn(\mathbb R)$ by taking the matrix $\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}$. Indeed it is invertible because its determinant is not zero but it can't be the square of a real matrix beacause its determinant is strictly negative.
Best Answer
Note that the image includes all real matrices $n\times n$ matrices that have $n$ distinct positive eigenvalues. The set of all such matrices is open and nonempty, and therefore has positive Lebesgue measure. Indeed, since this set is closed under multiplication by positive scalars, the Lebesgue measure must be infinite.