Your final statement is correct: any such relation must include all of the pairs $(a,a),(b,b)$, $(c,c),(d,d)$, and $(e,e)$ just in order to be reflexive, it must include $(a,b)$ to satisfy (iii), and once it contains $(a,b)$, it must contain $(b,a)$ in order to be symmetric. Call this relation $R_0$: $R_0=\{(a,a),(b,b),(c,c),(d,d),(e,e),(a,b),(b,a)\}\;.$ It is indeed special according to the definition, but it’s not the only special relation on $X$. For instance, we could add the pairs $(a,c)$ and $(c,a)$ to get the relation $\{(a,a),(b,b),(c,c),(d,d),(e,e),(a,b),(b,a),(a,c),(c,a)\}\;,$ which (as you can easily check) is also special. In fact you can add any pair to $R_0$ provided that you also add the reversed pair, so that you preserve the symmetry of the relation. In other words, if $\{x,y\}$ is any pair of distinct members of $X$ other than $\{a,b\}$, you can add the pairs $(x,y)$ and $(y,x)$ to $R_0$ and still have a special relation, but you have to add both of these pairs or neither.
How many pairs $\{x,y\}$ of distinct members of $X$ are there besides the pair $\{a,b\}$?
Let $P$ be the collection of all pairs of the kind that you counted in (1). If $S$ is any subset of $P$, you can add $(x,y)$ and $(y,x)$ to $R_0$ for every pair $\{x,y\}$ in $S$ and get a special relation on $X$, and every special relation on $X$ can be obtained in this way. This means that you can count special relations on $X$ by counting subsets of $P$. How many different subsets of $P$ are there? You’ll need your answer from (1) in order to answer this.