Let $\pi$ be a representation of the Banach *-algebra $A$ on $H$, prove that every nonzero $h\in H$ is cyclic for $\pi$ if and only if whenever $T\in B(H)$ commute with all of the $\pi(a),\ a\in A$ then $T$ is a scalar multiple of the identity.
I am struggling with The only if direction :
We can prove that every nonzero $h\in H$ is cyclic for $\pi$, which means $\overline{\{\pi(a)h|\ a\in A\}}= H$ is equivalent to the fact:
every closed subspace of $H$ invariant under all of the $\pi(a)$ is either $\{0\}$ or $H$
With this formulation we can obtain that $Ker(T-\lambda I)= H$ or $\{0\}$ for all $\lambda\in \mathbb{C}$, as $T-\lambda I$ commutes also with all of the $\pi(a)$, so we just need to argue why it is not possible that $Ker(T-\lambda I)= \{0\}$ for all values of $\lambda$ , in other words we need to find an eigenvector for $T$ , and it doesn't seem it should come from the commuting hypothesis.
There is no reason for any of the involved operators to have eigenvalues, so I don't see how your approach could be made to work.
Your argument should also use that $A$ is a $*$-algebra, as the result is not true if $A$ is not selfadjoint.
We use the common notation $$\pi(A)'=\{T\in B(H):\ T\pi(a)=\pi(a)T\ \text{ for all } a\in A\}.$$
Assume first that $\pi(A)$ is non-degenerate and the only invariant subspaces for $\pi(A)$ are $\{0\}$ and $H$. Since $\pi(A)$ is selfadjoint and non-degenerate, $\pi(A)'$ is a von Neumann algebra. In particular, it is generated by its projections. Let $P\in\pi(A)'$, and $K=PH$. For any $a\in A$, $$ \pi(a)K=\pi(a)PH=P\pi(a)H\subset PH=K. $$ So $K$ is invariant for $\pi(A)$, and thus either $K=\{0\}$ or $K=H$, which means either $P=0$ or $P=I$. Then $\pi(A)'=\mathbb CI$.
Conversely, suppose that $\pi(A)'=\mathbb C I$. Let $K\subset H$ be invariant for $\pi(A)$. From $\pi(a)K\subset K$, we get that if $P$ is the orthogonal projection onto $K$, then $$ \pi(a)P=P\pi(a)P,\qquad a\in A. $$ If $a\in A$ is selfadjoint, then $$ \pi(a)P=P\pi(a)P=(P\pi(a)P)^*=(\pi(a)P)^*=P\pi(a). $$ So $P$ commutes with the selfadjoint elements in $\pi(A)$. But $\pi(A)$ is a $*$-algebra, so every element is a linear combination of selfadjoints (namely, $a=(a+a^*)/2 + i (a-a^*)/(2i)$ ). So $P\in\pi(A)'$ is a scalar multiple of the identity, i.e. $P=0$ or $P=I$. That is, $K=0$ or $K=H$.