$\newcommand\SU{\text{SU}}$Vectorizing the matrix $u$ (by writing $a_i:=u_{k,l}$ for $i:=(k,l)$), we reduce the problem to the following: 

> For a complex vector $a=(a_i)$ with $\sum_i|a_i|^2$ being a known number $N>0$, we know the matrix $p=(p_{ij})$, where **$p_{ij}=\bar a_i a_j$**. Can we recover $a$ based on this information?

To answer this question, note that the condition $\sum_i|a_i|^2=N>0$ implies that $a_i\ne0$ for some $i$. Fix any such $i$. 

Then $p_{ii}=\bar a_i a_i\ne0$, and from the condition $p_{ij}=\bar a_i a_j$ we get $a_j=C_{ij}a_i$, where $C_{ij}:=p_{ij}/p_{ii}$ is a known complex number. Using the condition $\sum_j|a_j|^2=N>0$ again, we get 
$$0<N=\sum_j|a_j|^2=|a_i|^2 M_i,$$
where $M_i:=\sum_j|C_{ij}|^2$ is known. It also follows that $M_i>0$ and hence 
$$|a_i|=r_i:=\sqrt{N/M_i}$$
and $r_i$ is known. 

So, $a_i=r_i e^{it}$ for some $t\in[0,2\pi)$ and hence 
$$a_j=C_{ij}r_i e^{it}$$
for all $j$. 

Thus, we have determined all the $a_j$'s up to the constant factor $e^{it}$ of modulus $1$. 

Thus, we have determined the matrix $u$ up to the constant factor $e^{it}$ of modulus $1$. 

This is all that can be done. Indeed, if $u$ is a unitary matrix such that $u^*\otimes u$ is the given matrix $\SU$, then for any real $t$ we have that $v:=e^{it}u$ is a unitary matrix such that $v^*\otimes v=\SU$. $\quad\Box$