$\mathbf{Verification \ Requested:}$
$\mathbf{Attempt}:$ Let $f(z)= \displaystyle\sum_{n=0}^{\infty}a_nz^n$. Since $l$ is a non-isolated zero of $f(z)$, then we can find a sequence $\{x_p\}$ converging to $l$ such that $f(x_p)=0$ and $x_p \neq 0$ for all $p$.
Now, we put $l-y_p=x_p$. As $n \to \infty$, $y_p \to 0$ and $y_p \neq 0$ for all $p$.
Let us consider $f(l-z)=\displaystyle\sum_{n=0}^{\infty}a_n(l-z)^n=\displaystyle\sum_{n=0}^{\infty}b_nz^n=g(z)$.
Now, $g(0)=b_0=0$.
$\therefore g(z)=\displaystyle\sum_{n=1}^{\infty}b_nz^n$. Now $g(y_p)=\displaystyle\sum_{n=1}^{\infty}b_ny_p^n=y_p\displaystyle\sum_{n=1}^{\infty}b_ny_p^{n-1}=0 \implies \displaystyle\sum_{n=1}^{\infty}b_ny_p^{n-1}=0 $. Hence, $\displaystyle\lim_{p \to \infty} \displaystyle\sum_{n=1}^{\infty}b_ny_p^{n-1}=0 \implies b_1=0$
Suppose for $n=m$, $b_m =0$.
$\therefore$ $g(z)=\displaystyle\sum_{n=m+1}^{\infty}b_nz^{n-m} \implies g(y_p)=y_p\sum_{n=m+1}^{\infty}b_ny_p^{n-m-1}=0 \implies \sum_{n=m+1}^{\infty}b_ny_p^{n-m-1}=0$. Taking $p \to \infty$, we get $b_{m+1}=0$. Hence, $\forall n \in \mathbb{N}, \ b_n =0$.
$\therefore \ g(z)=0=f(l-z) \implies g(l-z)=0=f(z)$.
Is this correct?
Kindly verify.