I tried to prove the above mentioned statement written in the book "Serial Rings" by Gennadi Puninski.
For the first part I assumed that the module $E(M)$ is indecomposable. I was going by definitions of each term used.
A module $E(M)$ is indecomposable means $E(M)$ is non-zero and $E(M)$ can't be written as a direct sum of two non-zero submodules.
To prove $M$ uniform, we have to prove that every non-zero submodule of $M$ is essential, that is, every non-zero submodule of $M$ has a non-zero intersection with other submodules of $M$.
Now $E(M)$ is a module itself, which is the injective envelope of $M$, that is there exists an injective homomorphism $f: M \to E(M)$ and $f(M)$ is essential in $E(M)$.
Let us consider that $M$ is not uniform. Then $$M \cap T = \{0\} $$ for all other submodules $T$ of $M$.
Then $M$ is a zero-module.
I can't proceed further. Please help me.
As you started, suppose $E(M)$ is indecomposable.
Let $N$ be a nonzero submodule of $M$. Then $E(N)$ is a submodule of $E(M)$.
Case 1) If $E(N)=E(M)$, that would imply $N$ is essential in $E(M)$, hence it is also essential in $M$.
Case 2) If $E(N)\neq E(M)$, then it is a direct summand (since it is an injective module.) But that would mean that $E(M)=E(N)\oplus K$ for a nonzero $K$, and that $E(M)$ is decomposable. This is not possible in light of our hypothesis.
Therefore for any nonzero submodule $N$ of $M$, only Case 1 happens.
The other direction is about as straightforward: assume $M$ is uniform. If $E(M)=A\oplus B$, then $M=(M\cap A)\oplus (M\cap B)$. By definition of uniformity, it is going to be impossible to have $M\cap A$ and $M\cap B$ nonzero. In the case $M\cap A=\{0\}$, you may deduce $A=\{0\}$ because $M$ is essential in $E(M)$. Similarly if $M\cap B=\{0\}$ you may deduce $B=\{0\}$. So there you have it, $E(M)=A\oplus B$ implies one of $A$ or $B$ is already trivial.