How can I define $\mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set? Suppose in the axioms of $\sf ZF$ we replaced the Axiom of infinity 

There exists an inductive set. 

with the Axiom of Dedekind-infinite set

There exists a set equipollent with its proper subset.


How can I define the set of natural numbers $\mathbb{N}$ in this setting, and prove that it is the unique minimal inductive set?

I was informed by an expert that this answer is is 'practically that of Andrés' (see comment). Here we deploy less sophistication so it should be of interest to the student just starting to explore mathematical concepts.

We begin by defining,  up to isomorphism, the algebraic structure $(\Bbb N,+,*)$.
Let $f: A \to A$ be an injective mapping that is not surjective.
Let $a_0 \in A \setminus f(A)$.
Consider the statement
$\quad {\mathbf {S}: \displaystyle \exists \mathbf {I} \,\bigr(\,a_0 \in \mathbf {I} \,\land \,\forall x\in \mathbf {I} \,(x \in \text{Domain of } f \land f(x) \in \mathbf {I})\,\bigr)}$
(c.f. axiom of infinity)
Since
$\quad {\displaystyle a_0 \in A  \,\land \,\forall x\in A \,\bigr(x \in \text{Domain of } f \land f(x) \in A\bigr)}$
statement $\mathbf {S}$ is true.
The intersection of all sets $J$ satisfying
$\quad {\displaystyle a_0 \in J  \,\land \,\forall x\in J \,\bigr(x \in \text{Domain of } f \land f(x) \in J\bigr)}$
is our definition of the natural numbers $\Bbb N$ (here a subset of $A$). In essence, we're applying Peano's method in this axiomatic context. The object of interest is the function $f$ restricted to $\Bbb N$ giving us the abstract Peano successor function,
$\quad f = \sigma: \Bbb N \to \Bbb N$
The Peano construction can be found in Appendix I of Serge Lang's Undergraduate Algebra.
For the next step, use recursion to define a mapping $g$ defined on $\Bbb N$.
Set
$\quad g(0) = \emptyset$
With $g(n)$ defined, set
$\quad g(n+1) = g(n) \cup \{g(n)\}$
We can now construct the OP's minimal inductive set,
$\quad \displaystyle \omega = \bigcup_{n \in\Bbb N} g(n)$
