Let's answer this question in two ways (largely the same).
Firstly, we might consider the inverses of each element. In the Klein group, every element is its own inverse. In $\mathbb{Z}_4$, neither $1$ ($1 + 1 = 2$) nor $3$ ($3 + 3 = 2$) are their own inverses while $0$ and $2$ are. So they're not isomorphic.
Secondly, we might consider the subgroups of each. What are the subgroups of $\mathbb{Z}_4$? There are only 3 subgroups: the two trivial subgroups and the group $\langle 0, 2 \rangle$. (If $1$ is in the subgroup, then so are $1 + 1 = 2, 1 + 1 + 1= 3$, and $1 + 1 + 1 + 1 = 0$; if 3 is in the subgroup, then so are $3 + 3 = 2, 3 + 3 + 3 = 1, and 3 + 3 + 3 + 3 = 0$). What about the Klein group? There are at least 5 different subgroups: the two trivial groups, and the groups $\langle a,b \rangle$, $\langle a,c\rangle$, and $\langle a,d\rangle$ (where I assumed that $a$ was the identity element). So they're not isomorphic.
Another way equivalent to the second (and let's be honest - to the first as well) is to consider the orders of the elements. In the Klein group, there are 3 elements of order 2 and the identity. In $\mathbb{Z}_4$, there are 2 elements of order 4 ($1$ and $3$), an element of order 2 ($2$), and the identity. So they're not isomorphic.
One should note that the subgroups in the second point are exactly the cyclic subgroups generated by a particular element. Because there are different numbers of elements with the same order, there are different subgroups. One might be made to think that if two groups have the exact same number of elements of each order, than they must be isomorphic. But this is not true! The smallest counterexample are two groups of order 27. And that's pretty cool.