The answer to the question is no. If the sequence $\alpha = (\alpha_{n})_{\geqslant 1}$ satisfies the condition $c<\alpha_{n}<c^\prime$ there is no other condition on a sequence $\alpha$ for the answer to be affirmative. Let's assume that the answer to the question is affirmative. That is, there exists $\alpha = (\alpha_{n})_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for every open ball $B(x,\epsilon)\subset \ell^p({\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z\in \ell^p({\mathbb{N}})$ there exists $y=(y_{n})_{n\in \mathbb{N}} = (y_0,y_1,y_3,\ldots,y_n,\ldots)$ satisfying the condition
$$
{\rm (i)}\; y\in\ell^{p}(\mathbb{N}),
\qquad
{\rm (ii)}\;y\in L^{-K}_{\alpha}(z)
\qquad \mbox{ and } \qquad
{\rm (iii)}\; y\in B(x,\epsilon)
$$
We observe that for every integer $K\geqslant 1$ and every $({\bf u}_{n})_{n\in\mathbb{N}}\in \ell^p(\mathbb{N})$ we have, in a susceptible way,
$$
L^{K}_{\alpha}\Big( ({\bf u}_{n})_{n\in\mathbb{N}} \Big) = \left( \Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}\Big)\cdot {\bf u}_{K+n} \right)_{n\in\mathbb{N}}.
$$
From the observation we made above about the $L_{\alpha}$ operator the first condition below is an immediate consequence of the assumption we made. Furthermore, the second condition is a consequence of the first condition and the third condition is a consequence of the second condition.

**Condition 1.** there exists $\alpha = (\alpha_{n})_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p({ \mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z_{n})_{n\in \mathbb{N}}$ there exists $ y=(y_{n})_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \|y\|_{p}<\infty,
\qquad
{\rm (ii)}\;\Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}\Big)\cdot y_{K+n}=z_{n} \quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\;\| y-x\|_p<\epsilon
$$
**Condition 2.** there exists $\alpha = (\alpha_{n})_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p( {\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z_{n})_{n\in \mathbb{N}}$ there exists $ y=(y_{n})_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \sum_{n=0}^{\infty}|z_n|^{p}<\infty,
\qquad
{\rm (ii)}\; y_{K+n}= \Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot z_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\;\sum_{n=0}^{K-1}\left|y_{n} - x_n\right|^{p}
+
\sum_{n=K}^{\infty}\left|\Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot z_{n} - x_n\right|^{p} <\epsilon
$$
**Condition 3.** there exists $\alpha = (\alpha_{n})_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-\{0\}}$
such that for any open ball $B(x, \epsilon)\subset \ell^p( \mathbb{N})$
there exists a $K\in \mathbb{N}$ such that for every pair $z=(z_{n})_{n\in \mathbb{N}}$
and $\tilde{z}=(\tilde{z}_{n})_{n\in \mathbb{N}}$ in $\ell^p(\mathbb{N})$
there exists $ y=(y_{n})_{n\in \mathbb{N}}$ which fits both $z$ and $\tilde{z}$ satisfying

$$
{\rm (i)}\; \sum_{n=0}^{\infty}|z_n|^{p}<\infty,
\qquad
{\rm (ii)}\; y_{K+n}= \Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot z_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\sum_{n=0}^{K-1}\left|y_{n} - x_n\right|^{p}
+
\sum_{n=K}^{\infty}\left|\Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot z_{n} - x_n\right|^{p} <\epsilon
$$
$$
{\rm (iv)}\; \sum_{n=0}^{\infty}|\tilde{z}_n|^{p}<\infty,
\qquad
{\rm (v)}\; y_{K+n}= \Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot \tilde{z}_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (vi)}\; \sum_{n=0}^{K-1}\left|y_{n} - x_n\right|^{p}
+
\sum_{n=K}^{\infty}\left|\Big( \prod_{\ell=1}^{K}\alpha_{\ell+n}^{-1}\Big)\cdot \tilde{z}_{n} - x_n\right|^{p} <\epsilon
$$

Since $z=(z_{n})_{n\in \mathbb{N}}$ and $\tilde{z}=(\tilde{z}_{n})_{n\in \mathbb{N}}$ are arbitrary we can choose their $K$ first coordinates equal and the others different such that $\|\tilde{z}-z \|_p\geqslant 3\epsilon $. Adding member by member the equalities {\rm (ii)} and {\rm (v)} will result in $z=(z_{n})_{n\in \mathbb{N}}=\tilde{z}=(\tilde{z}_{n})_{n\in \mathbb{N}}$ since $0<c<\alpha_{n}$. However, this is a contradiction.