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).
I've come to a solution somewhat simpler.
Let D be a strongly connected digraph. Let $C = v_1 v_2 ... v_m$, $v_m = v_1$, be an odd cycle in the underlying graph. For each i, from 1 to m-1, let $W_i$ be a minimal walk from $v_i$ to $v_{i+1}$. If both vertices are connected by an edge $e_i = v_iv_{i+1}$, then $W_i = v_ie_iv_{i+1}$. Else, if the edge is $v_{i+1}v_i$, then suppose $W_i$ is an odd length walk (otherwise, we'd have an odd length cycle and be done by now). Since m-1 is odd, we have constructed an odd number of odd length walks. Concatenating them as $W_0 . W_1 . ... W_{m-1}$, we have a closed odd walk. Since every closed odd walk has a closed odd cycle, we're done.
Best Answer
Your SCCs are correct. I'm not sure how to parse your description of the condensation, but its edges should be $C_1\to C_2$, $C_2\to C_3$, $C_2\to C_4$ and $C_2\to C_5$.