First, a note on notation: $L(M)$ is a language which is a set. Languages don't accept or reject, Turing machines do. You can either say "$\epsilon \in L(M)$" or "$M$ accepts $\epsilon$".
There are many Turing machines in this language that don't meet your description. For example, imagine a machine that upon reading a blank goes and writes, say, 111 and then accepts. The initial state does not need to have a transition to an accept state on a blank symbol in order to belong in this language.
Here's another perspective on why this language is undecidable using Kleene’s Recursion Theorem. Imagine that there is some decider $D$ that, given the description of a TM $\langle M \rangle$, $D$ accepts if $M$ accepts $\epsilon$ and rejects otherwise. Now create a Turing machine with the following behavior:
Get my own description <M> and then run D on <M>.
If D accepts <M>:
If input is empty, reject. Otherwise accept.
If input is empty, accept. Otherwise reject.
The existence of this decider allows us to create the above Turing machine which is a contradiction. Essentially, if the decider says the Turing machine is supposed to accept the empty string, it then goes and rejects and if it's supposed to reject, it then goes and accepts. This idea of being able to use self-reference and ask "What am I about to do?" and then doing the opposite shows that no single machine can predict behavior of all others.