Theorem: Sets of Operations on Set of 3 Elements/Automorphism Group of B/Operations with Identity

:None of the operations of $\BB$ has an identity element.

Proof:
Recall Automorphism Group of $\BB$.
Consider each of the categories of $\BB$ induced by each of $a \circ a$, $a \circ b$ and $a \circ c$, illustrated by the partially-filled Cayley tables to which they give rise:
;$(1): \quad a \circ a$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a &   &   \\
b &   & b &   \\
c &   &   & c \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b &   &   \\
b &   & c &   \\
c &   &   & a \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c &   &   \\
b &   & a &   \\
c &   &   & b \\
\end {array}$
;$(2): \quad a \circ b$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & a &   \\
b &   &   & b \\
c & c &   &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & b &   \\
b &   &   & c \\
c & a &   &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & c &   \\
b &   &   & a \\
c & b &   &   \\
\end {array}$
;$(3): \quad a \circ c$
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & c \\
b & a &   &   \\
c &   & b &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & b \\
b & c &   &   \\
c &   & a &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & c \\
b & a &   &   \\
c &   & b &   \\
\end {array}$
The Cayley table of an operations on $\BB$ is constructed from combining one of the partial Cayley tables from each of the above categories.
It is then possible to identify which of these partial Cayley tables can contribute towards an operation with an identity element by seeing whether they contain at least one element of the nature:
:$x \circ y = x$
or:
:$x \circ y = y$
Of these, an operation with an identity element would need to combine:
The partial Cayley table induced by $a \circ a = a$:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a &   &   \\
b &   & b &   \\
c &   &   & c \\
\end {array}$
Either of the partial Cayley tables induced by $a \circ b = a$ or $a \circ b = b$:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & a &   \\
b &   &   & b \\
c & c &   &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & b &   \\
b &   &   & c \\
c & a &   &   \\
\end {array}$
Either of the partial Cayley tables induced by $a \circ c = a$ or $a \circ c = c$:
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & a \\
b & b &   &   \\
c &   & c &   \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & c \\
b & a &   &   \\
c &   & b &   \\
\end {array}$
Combining the partial Cayley table induced by $a \circ a = a$ with those of $a \circ b = a$ and $a \circ b = b$ in turn:
$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & a &   \\
b &   & b & b \\
c & c &   & c \\
\end {array} 
\qquad
\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a & b &   \\
b &   & b & c \\
c & a &   & c \\
\end {array}$
and there is no need to consider those induced by $a \circ c = a$ or $a \circ c = c$, as it is immediately apparent that neither of these partial Cayley tables can define an operation with an identity element.
{{qed}}