Let me elaborate on Tsemo Aristide's anwser that the direct sum of abelian groups should be replaced by the so-called free product for non-abelian groups.
When we want to compare a construction from the theory of abelian groups with the situation of general groups, we first need a way of comparing the class of abelian groups with the class of groups in general. The usual way of doing this is to use categories. A category is just a collection of objects, together with special 'maps' between them (note that I'm glossing over a whole lot of details here, but the Wikipedia page on categories has a lot of information). For instance, there is the category of groups (with group homomorphisms as maps), the category of $\mathbb{R}$-vector spaces (with linear maps), the category of sets (with no restriction on the maps), etc.
In our case, we want to compare the category of groups with the category of abelian groups, and somehow transfer the notion of a direct sum from the latter to the former. To do that, we first need to express the direct sum in the language of categories, i.e. just refering to objects (abelian groups) and maps (group homomorphisms), and in particular without refering to elements of the groups involved. That may seem close to impossible at first glance, but it can be done using the following trick.
First, we notice that if $(A_i)_{i\in I}$ is any collection of abeliqn groups, then for any other abelian group $B$ there is a very natural (set-theoretic) bijection
$$\textrm{Hom}(\oplus_{i \in I} A_i, B) \to \prod_{i \in I}\textrm{Hom}(A_i, B),$$
where $\oplus_{i \in I} A_i$ is the direct sum of the $A_i$ (you should at this point take the time to write down which map this is, and convince yourself that this map is indeed bijective). In words, this says that a homomorphism from $\oplus_{i \in I} A_i$ to $B$ and a collection of morphisms from every $A_i$ to $B$ are 'the same thing'. In fact, it turns out that the direct sum is unique with this property: if $A$ is any abelian group such that for any abelian group $B$ we get a natural bijection like above, then $A$ is isomorphic to $\oplus_{i \in I} A_i$. So this gives us precisely a characterisation of the direct product in terms of just the groups and the homomorphisms.
Now that we know what a direct sum looks like in the category of abelian groups, we can try doing the same for general groups. So given a collection $(G_i)_{i \in I}$ of not necessarily abelian groups, we want to find a group $G$ with the property that for any group $H$, there is a natural bijection
$$\textrm{Hom}(G, H) \to \prod_{i \in I}\textrm{Hom}(G_i, H).$$
Based on the abelian case, we might expect that this group $G$ should be the direct product of the $G_i$, or perhaps the subset of the direct product $G_i$ of sequences with only finitely many non-identity components, but unfortunately these groups do not satisfy the above condition (you should verify this by finding some explicit examples of sets $I$ and groups $G_i$ and $H$ for which we do not have a bijection like above). The group that does satisfy the condition is the one called the free product of the $G_i$. This group is formed by considering finite words, where the letters are taken from all the sets $G_i$, and we may simplify words by multiplying two adjecent letters if they come from the same $G_i$, and we may remove any identity elements from the words (see the wiki page for a more precise definition). It is a nice exercise to show that in the case that all the $G_i$ are abelian, the direct sum of the $G_i$ is isomorphic to the abelianization of the free product, so the free product is indeed a generalization of the direct sum.
So we see that the 'correct' translation of the direct sum concept to general groups leads to free products, instead of cartesian products or subsets thereof. In other words, there is not a problem with considering 'direct sums' of non-abelian groups per se, but to preserve the properties that the direct sum has in the abelian case, we need to consider the free product instead of a direct sum.
You seem to be assuming that if the multiplication tables are different the groups are not isomorphic, but that is not true. What is true is that if the tables are the same the groups are isomorphic. Your argument that there are at most $n^{n^2}$ tables is correct, which shows that there are no more nonisomorphic groups than that.
You can do much better by noting that the table has to have each element once in each column of the table. As there are $n!$ possible columns there are at most $n!^n\approx \left(\left(\frac ne\right)^n\right)^n(2\pi n)^{n/2}$ nonisomorphic groups. For $n=10$ this is $4\cdot 10^{65}$ instead of $10^{100}$ and for $n=100$ it is about $10^{15797}$ instead of $10^{20000}$. Of course, you can do much better by considering that each row must also have one of each element, but I don't see a quick way to quantify that.
Best Answer
Since the question asks for equality of operations rather than isomorphism: take $G$ to be any elementary abelian $2$-group. Then $xy^{-1} x = y$ so the operation $\delta_G$ gives you no information and, for example, there are already two such structures on the two-element set (given by choosing one or the other of the two elements to be the identity).