For $m>1$ or $n>1$ the answer is, ``no.'' The definition of rotationally invariant is if the distribution of $A$ is the same as the distribution of $U A V^T$ for every pair of matrices $U$ and $V$ which are non-random and orthogonal matrices on $\mathbb{R}^m$ and $\mathbb{R}^n$, respectively. If $n\geq 2$, take a 1-parameter family of orthogonal matrices $V(\theta) = (v_{ij}(\theta))_{i,j=1}^n$ for $\theta \in [0,2\pi)$ such that
$$
\left(\begin{matrix} v_{11}(\theta) & v_{12}(\theta) \\
v_{21}(\theta) & v_{22}(\theta) \end{matrix}\right)\,
=\,
\left(\begin{matrix} \cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \end{matrix}\right)
$$
and $v_{ij}(\theta) = \delta_{ij}$ if $i>2$ or $j>2$. Take $U$ to be the identity.
Let $B = U A V^T$. Then if $A$ has all entries either $+1$ or $-1$ then $b_{11}$ must have distribution equal to the distribution of $a_{11} \cos(\theta) + a_{12} \sin(\theta)$ as $\theta$ varies over $[0,2\pi)$ and $a_{11}$, $a_{12}$ take their values $\{+1,-1\}$ equally likely, IID. In particular, it is easy to see that $b_{11}$ has a continuous distribution with probability density function $f(x)=\pi^{-1} (2-x^2)^{-1/2} \mathbf{1}_{(-\sqrt{2},\sqrt{2})}(x)$, because $b_{11}$ can be written as $\sqrt{2} \sin(\theta+\phi)$ for some $\phi\in\{\pi/4,3\pi/4,5\pi/4,7\pi/4\}$ depending on $a_{11}$ and $a_{12}$. In particular, this is not discrete. It is not just $+1$ and $-1$ like one had for $a_{11}$.
The symmetry that is there for this ensemble is if $U$ is an $m\times m$ non-random permutation matrix, and if $V$ is an $n\times n$ non-random permutation matrix, then $A$ is distributed as $U A V^{T}$ because of permutation-invariance (exchangeability) of the IID product measure on $\{+1,-1\}$. There is undoubtedly a large literature on this ensemble within random matrix theory. For example, there is this paper of Tao and Vu https://arxiv.org/pdf/0903.0614.pdf
for the case $m=n$ with interest in the asymptotic behavior of the smallest singular value for large $n$.
