Both balls need to be in one metric space.
I come up with idea of metric space $X ={a,b,c}$
and $d(a,b) = d(a,c) = 2$, $d(b,c) = 1, d(x,x) = 0$.
Then i can say that $B(a,1)$ and $B(b,1)$ are non isometric. The problem is - $B(b,1)$ is obviously closed (by definition) but i am not sure about $B(a,1)$ ( dots with <=1 distance from a)
$B(a,1) = \{a\}$, and $B(b,1) = \{b,c\}$ (assuming you're taking the closed balls throughout, because anything else would be weird), so your example works.
I feel like you're rather over-thinking it, though: you can take any two metric spaces and make a new metric space including both by defining everything in one to be some fixed distance from everything in the other. Thus, a simple example would be to take your metric space to be $[-1,1] \cup \{\ast\}$, with $[-1,1]$ having the obvious metric, and $d(x,\ast) = 2$ for every $x \in [-1,1]$. Then $\bar{B}(0,1) = [-1,1]$, and $\bar{B}(\ast,1) = \{\ast\}$, and these are clearly not isometric.