Boltzmann superposition: assume that there is no stress when $t\le 0$, so the integral starts from $0$ and $\sigma(0^{-})=0$:

$$$$\gamma(t)=\int_0^t J(t-t^\prime)\dot{\sigma}(t^\prime)\mathrm{d}t^\prime$$$$

Performing Laplace Transform yields:

\begin{align} \hat{\gamma}(s)&=\hat{J}(s)\hat{\dot{\sigma}}(s)\\ &=\hat{J}(s)\left(s\hat{\sigma}(s)-\sigma(0^{-})\right) \\ &=\hat{J}(s)\left(s\hat{G}(s)\hat{\dot{\gamma}}(s)-\sigma(0^{-})\right)\\ &=\hat{J}(s)\left(s\hat{G}(s)(s\hat{\gamma}(s)-\gamma(0^-))-\sigma(0^{-})\right)\\ &=s^2\hat{J}(s)\hat{\gamma}(s)\hat{G}(s) \end{align}

Here we let $\hat{f}(s):=\mathcal{L}\lbrace f(t)\rbrace(s)$ be the Laplace transformation and since the stress starts at time $0$, $\gamma(0^-)$ and $\sigma(0^-)$ are simply $0$. The 2nd to 3rd step is derived from the convolution relation $\sigma(t)=\int_0^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$. Cancelling out the $\hat{\gamma}(s)$ and rearranging the last equation give

$$$$\frac{1}{s^2}=\hat{J}(s)\hat{G}(s)$$$$

The inverse Laplace of above equation gives the convolution relation

$$$$t=\int_0^{t}J(t-t^\prime)G(t^\prime)\mathrm{d}t^\prime$$$$

The last step is substituting $J(t)=J_e + t/\eta$ into above convolution relation and making $t\to\infty$:

\begin{aligned} \color{blue}{t} &=\int_0^{t}\left(J_e+\frac{t-t^\prime}{\eta}\right)G(t^\prime)\mathrm{d}t^\prime\\ &= J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime + {\color{blue}{\frac{t}{\eta}\int_0^\infty G(t^\prime)\mathrm{d}t^\prime}} - \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime \end{aligned}

Since $\eta:=\int_0^\infty G(t^\prime)\mathrm{d}t^\prime$, the blue terms cancelled out, which gives $$J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime = J_e\eta= \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime = \frac{1}{\eta} \int_0^\infty G(t^\prime)t^\prime\mathrm{d}t^\prime$$ Therefore we have

$$J_e = \frac{1}{\eta^2}\int_0^\infty tG(t)\mathrm{d}t$$

$Q.E.D.$