Why doesn't $z^n\cdot\left(\frac{a+b}{z}\right)^n = (a+b)^n$ always hold?

In complex numbers, $(ab)^c=a^cb^c$ does not necessarily hold.

$$((-1)(-1))^{1/2}=1\ne(-1)^{1/2}(-1)^{1/2}=i^2=-1.$$

It looks like Maple has trouble evaluating the left hand side of your equation. You can help Maple by asking it to expand (or simplify) the left-hand-side of your expression before using the is command from the assume facility.

eqn := z^n*((a+b)/z)^n = (a+b)^n;
assume(z::real,n::integer);

As we saw in your question, is(eqn); returns false but

is(simplify(eqn));

returns true. I suppose this arises from limitations of the Assume Facility; as referred to on the help page, this may be documented in [Corless, Monagan] (I was unable to find an online copy of it).

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Side note

If $n$ is any complex number, we cannot say anything in general (non-uniqueness) but applying the assumption that $(a+b)/z>0$ seems to be enough for the Assume Facility (with a helping hand from the simplify command):

restart;
eqn := z^n*((a+b)/z)^n = (a+b)^n;

assume((a+b)/z>1);
is(simplify(eqn));

returns true although $z^n$ is ambiguous for $z,n \in \mathbb C$. This seems to be a result of the way Maple defines the complex exponential to have a unique solution, see Section 5.1 The Complex Exponential Function (document can only be opened in Maple).

_{[Corless, Monagan] Corless, Robert, and Monagan, Michael. "Simplification and the Assume Facility." Maple Technical Newsletter, Vol. 1 No. 1. Birkhauser, 1994.}