The following identities hold for all real $x>0$. Your equality is the second identity at $x=1$.
$$
-x + \frac{\pi}{12}(6x^2-1) + x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0,\\
x + \frac{\pi}{12}(6x^2-1) - x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(1+ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0.
$$

My proofs are series manipulations, which I'll now sketch. To prove the first identity, consider the product expansion for the gamma function:
$$
\Gamma(z) = \frac{e^{-\gamma z}}{z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)^{-1}e^{z/n}.
$$
Take the logarithm of both sides; the RHS is now an infinite sum. Expand the $\log\left(1+\frac{z}{n}\right)$ term as power series about $z=0$, and switch the order of summation. Now we have:
$$
\log{\Gamma(z)} = -\gamma z - \log{z} + \sum_{k=2}^\infty \frac{(-1)^k}{k}\zeta(k)z^{k}
$$
The polygamma $\psi^{(-2)}(z)$ (by which I mean the Mathematica function $\texttt{PolyGamma[-2,z]}$) is the antiderivative of $\log{\Gamma(z)}$ with $\psi^{(-2)}(0)=0$. Integrating term-by-term, plugging in $z=ix$, and taking just the imaginary terms, we obtain
$$
\Im{\psi^{(-2)}(ix)} = -x\log{x} + x + \sum_{\ell=1}^\infty \frac{(-1)^\ell}{2\ell(2\ell+1)}\zeta(2\ell) x^{2\ell+1}.
$$
To obtain the appropriate series expansion for $\text{Li}_2(e^{-2\pi x})$, start with its second derivative
$$
\frac{d^2}{dx^2} \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{2\pi}{e^{2\pi x}-1} = x^{-1} + \sum_{k=1}^\infty \frac{B_k}{k!}(2\pi)^k x^{k-1}.
$$
Here, $B_k$ is the $k^{\rm th}$ Bernoulli number. After integrating twice, the series we get for $\text{Li}_2(e^{-2\pi x})$ is
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x + \sum_{k=1}^\infty \frac{B_k(2\pi)^k}{k(k+1)k!} x^{k+1}.
$$
The $\frac{\pi}{12}$ term comes from the fact that $\text{Li}_2(1) = \zeta(2) = \frac{\pi^2}{6}$. Using the relation between Bernoulli numbers and even zeta values, this series may be rewritten
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x - \frac{\pi}{2}x^2 - \sum_{\ell=1}^\infty \frac{(-1)^\ell}{\ell(2\ell+1)} \zeta(2\ell) x^{2\ell+1}.
$$
To obtain my first identity, just add the series for $2\Im{\psi^{(-2)}(ix)}$ and $\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x})$; everything cancels beyond the quadratic term. The second identity---and identities for every $m + ix$, $m \in \mathbb{Z}$---follow from the first via the formula
$$
\psi^{(-2)}(z+1) = \psi^{(-2)}(z) + \frac{1}{2}\log(2\pi) + z\log{z} - z,
$$
which is itself a corollary of the functional equation $\Gamma(z+1)=z\Gamma(z)$.

Taking more antiderivatives will give us identities involving $\Re\psi^{(-3)},\Im\psi^{(-4)},$ etc., but this isn't particularly interesting now that we know where they come from. All of joro's identities are now accounted for, I believe.

The real part $\Re\psi^{(-2)}(ix)$ or $\Re\psi^{(-2)}(1+ix)$ picks out the odd, rather than even, zeta values in my series expansion. Because these numbers, $\gamma,\zeta(3), \zeta(5), \zeta(7), \ldots$, remain so mysterious to number theorists, I expect that $\Re\psi^{(-2)}(1+i)=1.13063...$ *does not* have an closed-form expression in terms of "simpler" elementary/special function values, appropriately defined. I also expect that a proof of any such statement, even irrationality, is beyond the scope of current knowledge---but I would be glad to be corrected on either point.