The usual way to show isomorphy between groups is indeed to simply construct the isomorphism, i.e. construct a function $\varphi$ that is a bijection and morphism.
However sometimes you can resort to neater 'tricks', by considering actions of groups and constructing equivalent permutation representations. For instance to show that $Aut(\mathbf C_2\times \mathbf C_2) \cong \mathbf S_3 \cong \mathbf D_6$ one could notice that all three groups act on three elements. The first one actually consists of all permutations of the non-trivial elements of $C_2\times \mathbf C_2$, (1,0), (0,1) and (1,1), the second group is by definition the group of all permutations of $\{1,2,3\}$ and the third group acts on a triangle in the plane where once again it consists of all possible permutations of the three vertices of the triangle.
All three groups act faithful (only the identity acts trivially) therefore any bijection between these sets (the non-trivial elements of $\mathbf C_2\times \mathbf C_2$, the set $\{1,2,3\}$, the angles of a triangle) will naturally induce an isomorphism between these groups. It's not really an alternative to constructing the isomorphism ofcourse, it just tells you where to start looking for one.
Another strategy that I've seen at work when dealing with groups defined by presentations is to first construct an epimorphism (which is rather easy, one only has to verify that all relations still hold) and then show by considerations about the group (e.g. simplicity) that its kernel must be trivial.
However, it's often very hard to think of such things. Here's something I read: During the golden years of discovery of sporadic simple groups two groups of equal order were constructed and even though this was suspected and quite some non-trivial things about these groups had been proven, it remained unsure for quite another while whether or not they were actually equal. (It turned out they were. It would be interesting if someone had a reference of this fact, I forgot where I read it.)
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Here's how this may help to solve your particular problem. Let $\mathbf D_8$ act on a square with vertices numbered $1,2,3,4$. Note that indeed $a = (1\ 2\ 3\ 4)$ and $b = (2\ 4)$ correspond to valid isometries of the square, let's call them $R$ (for rotation) and $D$ (for reflection along the diagonal). This implies that the map $ \varphi \colon \langle R,D\rangle \to \langle a,b\rangle $ is actually an isomorphism: each element of either groups uniquely corresponds to some permutation of $\{1,2,3,4\}$ resp. the vertices of the square. Therefore it is sufficient to show that $\mathbf D_8$ is generated by $R$ and $D$ (which is well-known) and you're done.