Your identity isn't right. We can compute the zeta regularized value
for the two infinite products in your identity and in general,
$$\prod_{n=-\infty}^\infty (n+a)
\quad\ne\quad
a \prod_{n=1}^\infty (-n^2) \prod_{n=1}^\infty\left(1 - \frac{a^2}{n^2}\right)
\quad\ne\quad 2 i\sin(\pi a)$$
In certain sense, the $i$ in the last term comes from
$$\prod_{n=1}^\infty(-1)^n = (-1)^{-\frac12} = \pm i$$
In the RHS, which sign of $i$ you get depends on how you interpret $-n$ when you
throw it to the machine of zeta function regularization. It in turn depends on the imaginary part of $a$.
Another thing is the factorization
$$\prod_{n=-\infty}^\infty (n+a) \stackrel{?}{=} a \prod_{n=1}^\infty (-n^2) \prod_{n=1}^\infty\left(1 - \frac{a^2}{n^2}\right)\tag{*1}$$
is not $100\%$ legal. When you juggle infinite terms like this, it misses another phase factor which depends on $a$. Instead of $\displaystyle\;\prod_{n=1}^\infty(-n^2)\;$, let us just compute the zeta regularized value of LHS of $(*1)$ directly:
$$\prod_{n=-\infty}^\infty (n+a) = a\prod_{n=1}^{\infty}(-n+a)(n+a)\tag{*2}$$
Let's consider the case $\Im \alpha > 0$. Notice
$$0 < \arg(x + a) < \pi,\quad\forall x \in \mathbb{R}$$
We need to interpret $-n + a$ as $e^{\pi i}(n - a)$ when we throw it to the machine.
The zeta function corresponding to the infinite product $(*2)$ is given by:
$$Z(s) \stackrel{def}{=} a^{-s} + \sum_{n=1}^\infty \left[ \frac{1}{ e^{\pi i s}(n - a)^s} + \frac{1}{(n + a)^s}\right]
= a^{-s} + e^{-\pi i s} \zeta(s,1-a) + \zeta(s,1+a)
$$
where $\displaystyle\;\zeta(s,t) = \sum_{n=0}^\infty \frac{1}{(n+t)^s}\;$ is the
Hurwitz zeta function.
By definition, the zeta regularized value of the product $(*2)$ is
$$\left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta
\stackrel{def}{=}
e^{-Z'(0)}
= a \exp\left[\pi i \zeta(0,1-a) - \zeta'(0,1-a) - \zeta'(0,1+a)\right]$$
Recall following relations for Hurwitz zeta function and Gamma function:
$$\begin{align}
\zeta(0,t) &= \frac12 - t\\
\left.\frac{\partial}{\partial s}\zeta(s,t)\right|_{s=0}
&= \log\Gamma(t) - \frac12\log(2\pi)\\
\Gamma(1+a) &= a \Gamma(a)\\
\Gamma(a)\Gamma(1-a) &= \frac{\pi}{\sin\pi a}
\end{align}$$
We find when $\Im a > 0$,
$$
\left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta
= \frac{2\pi a e^{(a - \frac12) \pi i}}{\Gamma(1-a)\Gamma(1+a)}
= 2e^{(a - \frac12) \pi i}\sin\pi a
$$
If one repeat the argument for $\Im a < 0$, we need to interpret $-n + a$ as $e^{-\pi i}(n - a)$ and we will pick out a different phase factor $e^{-(a - \frac12) \pi i}$. Combine these, we have:
$$\left[ \prod_{n=-\infty}^\infty (n+a) \right]_\zeta
= 2e^{\epsilon(a - \frac12) \pi i}\sin(\pi a)
\quad\text{ where }\quad \epsilon = \begin{cases}
+1,& \Im a > 0\\
???,& \Im a = 0\\
-1,& \Im a < 0
\end{cases}$$
As you can see, this is in general different from the last term $2i\sin(\pi a)$ in your expression.
If we do the same thing to the product $\displaystyle\;\prod_{n=1}^\infty(-n^2)\;$ in RHS of $(*1)$, we find
$$\left[ \prod_{n=1}^\infty (-n^2) \right]_\zeta
= 2\pi e^{-\epsilon\frac{\pi}{2} i}
\quad\text{ when }\quad "-n" \text{ is interpreted as } e^{\epsilon \pi i} n.
$$
Together with the identity
$$\prod_{n=1}^\infty \left(1 - \frac{a^2}{n^2}\right) = \frac{\sin\pi a}{\pi a}$$
The RHS of $(*1)$ becomes $2 e^{-\epsilon\frac{\pi}{2} i}\sin(\pi a)$ and differs from
LHS of $(*1)$ by a phase factor $e^{\epsilon a \pi i}$.
