Atkinsons Theorem states that a bounded linear operator $T\in L(X,Y)$ where $X,Y$ are Banach spaces is a Fredholm operator if and only if there exist bounded linear operators $S,S'\in L(Y,X)$ and compact operators $K \in K(X,X)$ and $K'\in K(Y,Y)$ such that $$ST=Id_X+K,\quad TS'=Id_Y+K'.$$
I found that any closed operator so that S,S',K,K' exist that satisfy the above property on its domain is a Fredholm operator. Is the reverse also true?
Let $T$ be a closed unbounded Fredholm operator with domain $D$. Let show the existence of operators of the type you are searchign for.
First note that $\ker(T)$ and $R(T)$ are closed, then $$T':D/\ker(T)\to R(T),\qquad [x]\mapsto Tx$$ still has closed graph and is now bijective. Let $A:R(T)\to X/\ker(T)$ be the inverse composed with the inclusion of $D/\ker(T)$ to $X/\ker(T)$. Since $T'$ and $A$ have the same graph the graph of $A$ is also closed, by the closed graph theorem $A$ is bounded.
Now let $Y=R(T)\oplus F$ with $F$ finite dimensional, and $X=E\oplus\ker(T)$. From this decomposition you get a natural isomorphism $\iota:X/\ker(T)\to E$. Now define $$S:R(T)\oplus F\to E\oplus \ker(T),\qquad (r,f)\mapsto ( (\iota\circ A)(r), 0)$$ Then, with respect to these decompositions: $$(ST-\Bbb 1_X)\,(e,k)=(0,-k),\qquad (TS-\Bbb 1_Y)\,(r,f)=(0,-f).$$ and $ST-\Bbb 1_X$ is minus the projection to $\ker(T)$, while $TS-\Bbb 1_Y$ is minus the projection to $F$. These spaces are finite dimensional and so these maps are finite rank, hence compact.