Depending on what you mean by closed-form, the answer is to some extent, yes.
As per my comment, if $\Psi_1=\Psi_2$, then it can be shown that $A = A_1+A_2$ follows a Wishart distribution with scale matrix $\Psi_1$ and degrees of freedom $\nu_1+\nu_1$.
On the other, if $\Psi_1\neq \Psi_2$ (but they are both square and with $p$ rows), the distribution is not Whishart anymore. Indeed, in this case, the pdf of $A$ is
$
\left\{2^{\frac{1}{2}(\nu_1+\nu_2)p}\Gamma_p\left(\frac{1}{2}(\nu_1+\nu_2)\right)\det(\Psi_1)^{\frac{1}{2}\nu_1}\det(\Psi_2)^{\frac{1}{2}\nu_2}\right\}^{-1}\text{etr}\left(-\frac{1}{2}\Psi_1^{-1}A\right)\det(A)^{\frac{1}{2}(\nu_1+\nu_2-p-1)} {_1}F_1\left(\frac{1}{2}\nu_2;\frac{1}{2}(\nu_1+\nu_2);\frac{1}{2}(\Psi_1^{-1}-\Psi_2^{-1})A\right), A>0,
$
where $\Gamma_p(a)= \int_{A>0}\text{etr}(-A)\det(A)^{a-\frac{1}{3}(p+1)}dA$, where $\text{Re}(a)>\frac{1}{2}(p-1)$ and the integral is over the space of $p\times p$ symmetric positive definite matrices, ${_1}F_1$ is the confluent hypergeometric function (of the first kind) and $\text{etr}(A) = \exp(\text{tr}A)$.
For details and proofs check Chapter 3 of Gupta and Nagar (1999) Matrix Variate Distributions, Monographs and Surveys in Pure and Applied Mathematics 104, Chapman & Hall/CRC.