eulerian-pathgraph theory
A directed graph has an Eulerian cycle if and only if every vertex has equal in degree and out degree, and all of its vertices with nonzero degree belong to a single strongly connected component.
I was looking for this proof in the Internet, but all i have found are the proofs for undirected graphs. I would be glad, if someone can give a link or name of the book, where i can find proof for this claim.
If you can find Hamiltonian cycles in an undirected graph, you can find them in directed graphs too by replacing each vertex by a linear sequence of three vertices *---*---* and connect all incoming edges to one of the end nodes and all the outgoing edges to the other end node. After that you don't need to remember the direction of the edges.
For an Eulerian circuit, you need that every vertex has equal indegree and outdegree, and also that the graph is finite and connected and has at least one edge. Then you should be able to show that
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).
Best Answer
Pages 5-8 of Sascha Wolff’s Lecture #1 “Round Trips” from his last-year block-course “Algorithmic Graph Theory” contain an illustrated proof both for undirected and directed case.