A loop is commonly defined as an edge (or directed edge in the case of a digraph) with both ends as the same vertex. (For example from $a$ to itself). Although loops are cycles, not all cycles are loops. In fact, none of the above digraphs have any loops.
Cycles are usually defined as closed walks which do not repeat edges or vertices except for the starting and ending vertex. This definition usually allows for cycles of length one (loops) and cycles of length two (parallel edges).
Note that cycles (and walks) do not make any reference to the orientation of the edges in question. Directed cycles (and directed walks) may only travel along the "forward" direction of the edges. In particular, that implies that $G_3$ pictured above has a third cycle, $(\color{blue}{(a,b)},(b,c),(c,a))$ where the $\color{blue}{(a,b)}$ refers instead to the edge pointing from $b$ to $a$.
Technically, all of the graphs above except for $G_2$ are directed multigraphs since in each you have parallel edges. Although in simple graphs (graphs with no loops or parallel edges) all cycles will have length at least $3$, a cycle in a multigraph can be of shorter length. Usually in multigraphs, we prefer to give edges specific labels so we may refer to them without ambiguity.
As for being strongly connected, yes all of them are and your definition is correct.
Your additional question, "what is the difference between a cycle and a connected component"
The above graph contains a cycle (though not a directed cycle) yet is not strongly connected.
One can prove that if a directed multigraph is strongly connected then it contains a cycle (take a directed walk from a vertex $v$ to $u$, then a directed walk from $u$ to $v$. Any closed walk contains a cycle).
One can also show that if you have a directed cycle, it will be a part of a strongly connected component (though it will not necessarily be the whole component, nor will the entire graph necessarily be strongly connected).
A more general approach to this is given by topological sorting. In particular, a topological sort exists if and only if the graph is a directed acyclic graph. The 'algorithms' described in the other answers effectively perform a topological sort, in that they repeatedly remove vertices with no incoming/outgoing edges.
For instance, a topological ordering of your graph is given by $A,B,C,D,E,F$ and if you redraw the graph so that the earlier vertices are higher than later vertices (not necessarily linearly) you can see that there are no 'upward' edges, so there can be no cycle.
Best Answer
You have successfully found the only edge that works, but we can do this more systematically.
We can describe the graph we have right now in three groups:
If we add a new edge and get a cycle through all $8$ vertices, that cycle cannot contain all four edges $A \to B, B \to F, F \to E, E \to A$: then it would come back to $A$ too early. Similarly, it cannot contain all four edges $C \to D, D \to H, H \to G, G \to C$.
But the cycle must have $8$ edges; the only way to get to $8$ is to use $3$ edges from group 1, $3$ edges from group 2, the edge $F \to G$, and the new edge we're going to add.
If the cycle is using edge $F \to G$, then it can't use $F \to E$ or $H \to G$. (The cycles should only leave $F$ once, and it should only enter $G$ once.) This means we know exactly which edges of the current graph it uses.
Those edges can be put together into a $7$-edge path: $$E \to A, A \to B, B \to F, F \to G, G \to C, C \to D, D\to H.$$ The only edge that can turn this $7$-edge path into a cycle is the edge $H \to E$, which is the edge you found.
(The condition that the cycle must start at $A$ is a red herring: once we have a cycle through all $8$ vertices, we can start it anywhere we like.)