Cayley table with the identity along a diagonal
The "main diagonal" of a table/matrix refers to the entries with index $(i,i)$; this is the diagonal that runs from "top left" to "bottom right": it's the entries marked with a $\mathbf{D}$ below: $$\begin{array}{|c|c|c|c|} \hline \mathbf{D}& \cdot & \cdot & \cdot\\ \hline \cdot & \mathbf{D} & \cdot & \cdot\\ \hline \cdot & \cdot & \mathbf{D} & \cdot\\ \hline \cdot & \cdot & \cdot & \mathbf{D}\\ \hline \end{array}$$
If you have a Cayley table, and the elements along the top are ordered the same way as the elements along the side (so that the main diagonal entries correspond to $a*a$ for every $a$, and you already know that this is the Cayley table for a group, and every main diagonal entry is equal, then that entry must be the identity (since $e*e=e$ holds, so every entry must be $e$).
A group in which every element is its own inverse must be abelian: if $xx=e$ for every $x$, and $a$ and $b$ are any two elements, then we have that $(a*b)^2 = e = e*e = a^2*b^2$. So then we have $$a*b*a*b = a*a*b*b$$ and multiplying on the left by $a$ and on the right by $b$ we get $$b*a = a*b,$$ so the group is abelian.
To see an example of a group where this happens, consider the following operation: flipping a rectangular mattress. You can flip it end-to-end; you can rotate it without changing what is on top; you can flip it and rotate it; or you can do nothing. Each of these is an element of a group; if you do the same thing twice in succession, they cancel out. So every element of the group is its own inverse. There is absolutely no problem with that: there are lots of groups like that.
Here is what they look like:
Let us use $E$ to denote "Even", and $D$ to denote "Odd". We make a group with underlying set $\{E,D\}$, and add as follows: $E+E = E$; $E+D=D+E=D$; and $D+D=E$. (Even plus even and odd plus odd are both even; even plus odd and odd plus even or both odd).
Now let $n\geq 1$, and let $G$ be the set of all $n$-tuples $(x_1,x_2,\ldots,x_n)$, where $x_i$ is either $E$ or $D$. We add tuples by adding component by component: $$(x_1,x_2,\ldots,x_n) + (y_1,y_2,\ldots,y_n) = (x_1+y_1,x_2+y_2,\ldots,x_n+y_n),$$ where in each component we are adding following the rules of the previous paragraph. This is a group (it has $2^n$ elements); the identity element of the group is the element $(E,E,E,\ldots,E)$. And in this group, every element is its own inverse: $(x_1,\ldots,x_n) + (x_1,\ldots,x_n) = (E,E,E,\ldots,E)$, no matter what $x_i$ is: if $x_i=D$, then $x_i+x_i = D+D=E$; if $x_i=E$, then $x_i+x_i = E+E=E$. So either way, we get the identity.
However, if you simply happen to find a "Cayley table" on the ground in which every main diagonal entry is equal, this does not suffice to tell you that you have an abelian group before you: associativity is not easy to determine from just staring at the table, and it is possible to write down the Cayley table of a binary operation in which (i) every element appears exactly one in each row and in each column; (ii) the rows correspond to elements in the same order as the columns do; (iii) all the main diagonal entries are the same; but (iv) the table does not correspond to an associative operation (that is, you don't have a group).
Here's an example where this happens: $$\begin{array}{cc|c|c|c|} * && a & b & c\\ &&&&\\ \hline a && a & b & c\\ \hline b && c & a & b\\ \hline c && b & c & a\\ \hline \end{array}$$ Every element appears exactly once in each row and each column, but this table does not yield a group, because the operation $*$ it defines is not associative: $(b*c)*c = b*c = b$, but $b*(c*c) = b*a = c$.