Does Eulerian cycle in digraph really need strongly connected component?

I think this question is best clarified with the following observation:

Once we assume that in-degrees and out-degrees of every vertex are equal, weakly connected and strongly connected components become one and the same.

So it is correct to say that an Eulerian cycle in a digraph requires the graph to be strongly connected, but it is also correct to say that being weakly connected is enough.

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Here is a proof of the observation. Suppose we have a digraph $D$ which is weakly connected, but not strongly connected. Then $D$ can be decomposed into two parts, $D_1$ and $D_2$, which are connected by some $k>0$ edges oriented from $D_1$ and $D_2$, but with no edges oriented from $D_2$ and $D_1$.

If there are $m$ edges internal to $D_1$, then the sum of the out-degrees of all the vertices of $D_1$ is $m+k$, but the sum of the in-degrees of all the vertices of $D_1$ is only $m$. These are not equal, so there must be some vertex of $D_1$ where the out-degree is not equal to the in-degree. (In fact, there must be some vertex of $D_1$ where the out-degree exceeds the in-degree.)

Conversely, if out-degrees and in-degrees are equal at each vertex, then such a decomposition cannot exist: if $D$ is weakly connected, it must be strongly connected.

(More generally, if $D$ is not necessarily connected, but satisfies the degree condition at each vertex, then this argument applies to make each of its weak components strong.)