$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No, there is no uniqueness here.
Indeed, let $\hat f$ denote the Fourier transform of a (say integrable) function $f\colon\R\to\C$, so that $\hat f(t)=\int_\R e^{itx}f(x)\,dx$ for real $t$. Then $\widehat{u^*}(t)=\hat u(-t)^*$ for real $t$, and the equation $$u*u^*=A \tag{1}\label{1}$$ for $u$ becomes the equation $$\hat u(t) \hat u(-t)^*=\hat A(t) \tag{2}\label{2}$$ for $\hat u$.
Clearly, equation \eqref{2} can have multiple solutions. So, equation \eqref{1} can have multiple solutions.
E.g., suppose that $\hat A(t)=t^2e^{-t^2}$, $$\hat u_1(t)=te^{-t^2/2},\quad\text{and}\quad \hat u_2(t)=ite^{-t^2/2}(1(t>0)-1(t<0))$$ for real $t$. Then $\hat u_1$ and $\hat u_2$ are two different solutions of \eqref{2}.