None of them is incorrect; they are all correct within their magisteria. That's why we talk about "group isomorphism", "order isomorphism", "field isomorphism", etc. We omit the specific kind when it is understood from context. Asking which is "correct" and which is "incorrect" is like asking which of the many "Jones" in the telephone directory is "the real Jones", and which ones are impostors. And just like a particular surname may be common, if in your place of work there is only one person with that surname, they may be refered to as "Jones" without fear of confusion.
When you are working with groups, you are interested in group homomorphisms and group isomorphisms. When you are working with semigroups, you are interested in semigroup homomorphisms and semigroup isomorphisms. When you are working with ordered sets, yo uare interested in order homomorphisms and order isomorphisms. Etc.
The reason that an isomorphism corresponds to the "essentially the same object" idea is that a bijection works like a "relabeling" of the objects. Consider the act of translating numbers from English to Spanish. Addition of numbers should not depend on which language we are speaking, in the sense that since "two" corresponds to dos, and "three" corresponds to tres, we should expect "five" (which is "two" plus "three") to corrrespond to whatever dos mas tres (namely, cinco) corresponds. The properties that the numbers and addition of numbers have do not depend on what we call the numbers, but rather on the properties of the numbers. So, numbers-under-addition is "essentially the same, except for the names we use" as números-bajo-suma. An isomorphism is the way of saying that the only differences between the two objects, as far as the particular structure we are interested about is concerned are the "names" we give to the objects and the operations.
Your example deals with a very specific construction of $\mathbb{Z}$ in terms of $\mathbb{N}$. The identification is in fact an identification that carries a lot of properties; it is a bijection that respects (i) order; (ii) addition; (iii) multiplication; and (iv) any derived operation from these. There are other ways of defining $\mathbb{Z}$ that do include $\mathbb{N}$ as a subset. The point of the isomorphism is that it does not matter how we construct $\mathbb{Z}$ and how we construct $\mathbb{N}$, in the end we have a set with certain properties, sitting inside another set that has further properties, and these properties are maintained regardless of how we constructed these objects.
Correction: It is not quite correct that "homomorphism is the same as isomorphism but not necessarily a bijection". For example, in the case of "order homomorphism", the condition only requires that $a\leq b\implies f(a)\preceq f(b)$, and the converse is not required (though the converse is required for isomorphisms). It is a bit better to say that an isomorphism of partially ordered sets is a homomorphism that has an inverse that is also a homomorphism; this gives you the biconditional as a consequence, rather than as a premise, and it makes it fit into the general scheme better. (We do the same thing with topological spaces, where the concept of isomorphism is that of homeomorphism, which is a continuous map that has an inverse, and the inverse is also continuous. The definition works in the context of groups, rings, semigroups, fields, etc., but it turns out that in those cases, being a bijective homomorphism suffices to guarantee that the set-theoretic inverse is also a homomorphism.)
As to homomorphisms vs. isomorphisms: There is a very fruitful philosophy that is that maps between objects are more important than objects. If you look at vector spaces, you can stare at particular vector spaces and get a lot of nice things, but vector spaces don't truly come into their own (in terms of power, applicability, usefulness, etc) until you introduce linear transformations (which are homomorphisms of vector spaces). Groups are ubiquitous, but it is homomorphisms between groups (which allow you to consider representations, group actions, and many other things) that make them impressively useful. The real numbers, as a metric space, is very nice; but it is continuous functions (homomorphisms of metric spaces/topological spaces) that are the cornerstone of their applicability to physics and other contexts. Homomorphism are "functions that play nice with the structure we are interested in." Functions are very useful, the structure may be very useful, and homomorphisms is a way of getting the best of both worlds: functions, and structure.
Homomorphisms are more powerful than simply isomorphisms because isomorphisms don't really let us "change" the structure, it only let us change the names we give to the objects in the structure. It is homomorphisms that truly allow us to switch contexts. If you only allowed continuous bijections from $\mathbb{R}$ to $\mathbb{R}$ with continuous inverses (what an isomorphism of metric spaces is), you would get a lot of interesting stuff, but nowhere near what you get when you allow yourself to consider all continuous functions.
This philosophy (concentrate on maps between objects, not on objects themselves) is at the very core of Category Theory (as Pete Clark mentions in the comments), and of modern takes on many subjects.
Added. As Jackson Walters points out, I've downplayed the importance of isomorphisms above. One of the fundamental problems in almost every area of mathematics is "When are two objects that may look different actually the same?" This is usually stated in terms of a "classification problem," and it usually comes down to asking whether there is an easy way to tell if there is an isomorphism between two given objects without having to explicitly look for one, or whether there is some "tractable" complete list of all objects of interest up to isomorphism. Examples of the former are the theorem that says that two vector spaces over the same field are isomorphic if and only if they have the same dimension. Examples of the latter are the "Fundamental Theorem of Finitely Generated Abelian Groups" (that tells you that every finitely generated abelian group is isomorphic to one with a particularly nice structure), and to a lesser extent the classification of finite simple groups (which tells you that there are some families plus a finite list that contains all finite simple groups). Isomorphisms are important, especially as a way of simplifying the study of objects by helping us look at what the things "are" instead of what they look like.
Best Answer
The "structure" preserved is how the elements relate to each other under the operations of the algebraic structure. For example, for commutative rings, if $\ a = b^2 - c^2 = (b-c)(b+c)\ $ is a difference of squares then applying a ring hom $\,h\, :\, r\mapsto \bar r\,$ shows that it remains a difference of squares in the image ring, viz. $\ \bar a = \bar b^2-\bar c^2 = (\bar b - \bar c)(\bar b +\bar c),\,$ so this particular ring-theoretic structure is preserved. Similarly preserved are polynomial expressions, i.e. expressions composed of basic ring operations (addition, multiplication) and constants $\,0,1.\,$
For example, a root of a polynomial remains a root of the image of the polynomial. So we can prove that a polynomial has no roots by showing it has no roots in a simpler homomorphic image, e.g. a ring with size so small that we can easily test all of its elements to see if they are roots, e.g. common parity arguments: $\,f(x) = x(x\!+\!1)+2n\!+\!1\,$ has no integer roots since $\,x(x\!+\!1)\,$ is even, so adding $2n\!+\!1$ yields an odd hence $\ne 0;\,$ equivalently $\!\bmod 2\!:\ f(0)\equiv 1\equiv f(1),\,$ i.e. $\,f\,$ has no roots mod $\,2,\,$ hence it has no integer roots. This idea generalizes as follows
Parity Root Test $\ $ A polynomial $\,f(x)\,$ with integer coefficients has no integer roots when its $\rm\,\color{#0a0}{constant\,\ coefficient}\,$ and $\,\rm\color{#c00}{coefficient\,\ sum}\,$ are both odd.
Proof $\ $ If so then $\ \color{#0a0}{f(0)} \equiv 1\equiv \color{#c00}{f(1)}\,\pmod 2,\ $ i.e. $\:f\:$ has no roots in $\,\Bbb Z/2 = $ integers mod $\,2,\,$ therefore $\,f\,$ has no integer roots. $\ $ QED
In the same way, we can often reduce problems to "smaller", simpler problems in modular images. Because homs preserve the ambient algebraic structure, as above, we can often deduce information about the original problems from information gleaned in the simpler modular images. Such problem solving by modular reduction is an algebraic way of "dividing and conquering".