If $G$ is a finite group whose irreducible characters have degrees 1,1,4,4,5,5,6 then $G$ is isomorphic to $S_5$.
From the character degrees we immediately get that $G/[G,G] \cong C_2$ since there are exactly two irreducible characters of degree 1, and that $|G|=1^2 + 1^2 + 4^2 + 4^2 + 5^2 +5^2 +6^2 = 120$. We also know that such a group has only 7 conjugacy classes.
If you have classified the groups of order 120, then you know there are only 3 groups with $G/[G,G] \cong C_2$ and only one of those has 7 conjugacy classes, namely $G=S_5$.
If not, you can already extract a lot of information. From the character degrees we know that $G$ is not of the form $H \times C_2$ (only one copy of $6$). We know the index of the Sylow 5-subgroup is either 1 or 6, but if it is 1, then the character degree 5 is not possible (standard result from Isaacs's textbook, 6.15), so we get a group with 6 Sylow 5-subgroups. We know the focal subgroup for $p=2$ is index 2, and consulting our table of groups of order 8, we get the Sylow 2-subgroup is $C_2 \times C_2 \times C_2$ with a direct factor (no!) or $D_8$ with PGL fusion ($S_5$ is PGL(2,5)). So we already get the 2-local and 5-local structure.
In general though the character degrees don't have to tell you much about the group. What they do tell you about the group is an area of active research. The state of affairs in the late 1960s is summarized in chapter 12 Isaacs's textbook, and I believe many of his more recent papers have more up to date summaries.
I think your approach of studying the order of the elements is the right one. If you want to do that in a more systematic way, you may study order of elements in a direct product. Namely, say we have $G_1, \ldots, G_r$ some groups, some element $g = (g_1, \ldots g_r) \in G_1 \times \ldots \times G_r$ in the direct product, and we want to know the order of $g$ inside $G_1 \times \ldots \times G_r$. It turns out that if $n_i$ denotes the order $g_i \in G_i$, then the order of $g$ is $\operatorname{lcm}(n_1,\ldots,n_r)$. In particular, since the order $g_i$ divides $|G_i|$ (the number of elements of the group), then order of $g$ divides $\operatorname{lcm}(|G_1|,\ldots,|G_r|)$.
In your example, take $G_1 = \mathbb{Z}_2$ and $G_2 = \mathbb{Z}_4$. We know the order of any element inside $G_1 \times G_2$ divides $\operatorname{lcm}(|G_1|,|G_2|) = \operatorname{lcm}(2,4) = 4$. So there is no element of order $8$.
Using the same principle we may see for example that $ \mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic. Indeed, we know $(1,1)$ has order $\operatorname{lcm}(2,3)= 6$.
Best Answer
To make my comments into an answer . . .
A presentation for $\Bbb Z_m\times\Bbb Z_n$ is
$$\langle a,b\mid a^m, b^n, ab=ba\rangle.$$
One can think of $a$ as $([1]_m, [0]_n)$ and $b$ as $([0]_m, [1]_n)$.
A good rule of thumb for determining whether you're done is to show, whenever possible, that any element of the group can be derived from the candidate generators & relators, and that no other elements can be made; however, it might not be possible to define an algorithm here, since this is combinatorial-group-theory, a place where many things are undecidable (in the sense that no Turing machine exists such that, given a presentation, it will halt in finite time whether or not you're done - but there's so many such results that it's difficult to keep track, so don't take my word for it here; this is why I was hesitant about answering rather than commenting).