Generally these tables are interpreted as taking an $x$ from the left column and a $y$ from the top row and putting its product $xy$ in the $(x,y)$ position.
You have already been told that $d$ goes in the $(a,b)$ position, and that $e$ goes in the $(c,a)$ position, and that $b$ goes in the $(d,c)$ position. Since groups of prime order are abelian, you can also conclude what $ba,cd$ and $ac$ are.
Since $a,b,c,d,e$ are likely assumed to be distinct, you can also tell from these that the only candidate for the identity is $e$, and so that allows you to fill in the last column and the last row rapidly. At this point you also learn that $a$ and $c$ are inverses of each other.
Using that relationship, you can deduce from $ab=d$ that $b=cab=cd$, so another entry appears in the $(c,d)$ position. As you get further along, you should be able to deduce each position.
Don't forget also that you have another tool at your disposal, namely that all the elements satisfy $x^5=e$. Another thing is that $a,c$ are paired up as inverses, and $e$ is its own inverse... what can you conclude about $b$ and $d$? Also, show that $a^2\in\{b,d\}$: if you try both of them out, you should see immediately that only one is consistent with the relations.
Please update us with your progress.
Computer search finds these three completions and no others:
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & c & c & a \\
b & d & b & b & d \\
c & a & c & c & a \\
d & d & b & b & d \\
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & a & c & c \\
b & b & b & d & d \\
c & c & c & a & a \\
d & d & d & b & b \\
\end{array}
$$
$$\begin{array}{c|cccc}
\cdot & a & b & c & d \\\hline
a & a & b & c & d \\
b & a & b & c & d \\
c & d & c & b & a \\
d & d & c & b & a \\
\end{array}
$$
Best Answer
As Gerry Myerson said, first identify the identity element; you have enough information to do this. Identifying it will let you fill in its row and it column in the table. Then use the fact that up to isomorphism there are only two groups of order $4$, the cyclic group of order $4$ and the Klein $4$-group; the fact that $d^2=a$ is important for deciding which one you have.