 Solution

For the first Cayley table, associativity fails: $(a\ast a)\ast b=1\ast b=b$, whereas $a\ast (a\ast b)=a\ast c=d$. It therefore does not define a group.

For the second table, assign the residues $0,1,2,3,4$ to $1,a,c,b,d$, respectively. Every table entry agrees with addition modulo an integer modulo five. Thus this table defines the cyclic group of order five, with identity $1$ and inverses $a^{-1}=d$, $c^{-1}=b$. The identification with a known group verifies associativity as well as closure and the inverse axioms; inspecting only the identity row and inverses would not suffice.