Due to the bounty I'll give a second answer.

Let $r = \sqrt{-8\log \tau}$. The proposed asymptotic form of $$ f(x)= - \int_0^1 \mathrm d\tau \,x r \, J_1(x \tau r),\tag{1} $$ for large $x$ from my first answer  (note that the $O(x^{-1/2})$ term was too optimistic), $$ f_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) + c + o(1),\tag{2} $$ can be derived and refined in the following way: Obviously $f_\infty(x)$ asymptotically fulfills the ODE (see also the comment of @TheSimpliFire) $$ f_\infty(x)+\frac e 4 f_\infty''(x) \simeq c,\tag{3} $$ therefore we check if $f(x)$ also solves (3) for large $x$. We get \begin{align} F(x) &= f(x)+\frac e 4 f''(x) \tag{4a}\\ &=\int_0^1 \mathrm d\tau\, \frac r 4\big[x(e r^2t^2-4) J_1(x\tau r)-e r t J_0(x\tau r)\big]\tag{4b}\\ &\simeq\int_0^1 \mathrm d\tau\, \frac {xr} 4(e r^2t^2-4) J_1(x\tau r).\tag{4c} \end{align} Note that for large $x$ the integrand oscillates and decays rapidly with increasing $x$, such that the relevant contributions come from the lower bound, where $\tau\ll 1$. If the now substitute $\tau\to y/x$ and use $-\log \tau \to \log x-\log y \simeq \log x$, we get \begin{align} F(x)&\simeq-\sqrt{8\log x}\int_0^\infty \mathrm dy\,J_1\big(\sqrt{8\log x}\,y\big) =-1,\tag{5} \end{align} such that asymptotically $c=-1$. Note that the convergence is extremely slow. In conclusion, the asymptotics becomes $$ f_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) - 1 + o(1),\tag{6} $$ the constant $c=-1$ can be seen as the missing contribution from the $n=0$ term in the original sum (if we define $0^0=1$).

The next terms in the expansions seem to be $$ f_\infty(x) \simeq \sqrt 2 \cos(2 e^{-1/2} x) - 1 + d x^{-\epsilon} + e x^{-1}\sin(2 e^{-1/2} x),\tag{7} $$ with $d\approx -0.18$, $\epsilon\approx 0.17$, $e\approx -0.097$.

Due to the bounty I'll give a second answer.

Let $r = \sqrt{-8\log \tau}$. The proposed asymptotic form of $$ f(x)= - \int_0^1 \mathrm d\tau \,x r \, J_1(x \tau r),\tag{1} $$ for large $x$ from my first answer, $$ f_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) + c + o(1),\tag{2} $$ (note that the $O(x^{-1/2})$ term was too optimistic) can be derived and refined in the following way: Obviously $f_\infty(x)$ asymptotically ($\simeq$) fulfills the ODE (see also the comment of @TheSimpliFire) $$ f_\infty(x)+\frac e 4 f_\infty''(x) \simeq c,\tag{3} $$ therefore we check if $f(x)$ also solves (3) for large $x$. We get \begin{align} F(x) &= f(x)+\frac e 4 f''(x) \tag{4a}\\ &=\int_0^1 \mathrm d\tau\, \frac r 4\big[x(e r^2t^2-4) J_1(x\tau r)-e r t J_0(x\tau r)\big]\tag{4b}\\ &\simeq\int_0^1 \mathrm d\tau\, \frac {xr} 4(e r^2t^2-4) J_1(x\tau r).\tag{4c} \end{align} Note that for large $x$ the integrand oscillates and decays rapidly with increasing $x$, such that the relevant contributions come from the lower bound, where $\tau\ll 1$. If we now substitute $\tau\to y/x$ and use $-\log \tau \to \log x-\log y \simeq \log x$, we get \begin{align} F(x)&\simeq-\sqrt{8\log x}\int_0^\infty \mathrm dy\,J_1\big(\sqrt{8\log x}\,y\big) =-1,\tag{5} \end{align} such that asymptotically $c=-1$. Note that the convergence is extremely slow. In conclusion, the asymptotics becomes $$ f_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) - 1 + o(1),\tag{6} $$ the constant $c=-1$ can be seen as the missing contribution from the $n=0$ term in the original sum (if we define $0^0=1$).

The next terms in the expansions seem to be $$ f_\infty(x) \simeq \sqrt 2 \cos(2 e^{-1/2} x) - 1 + d x^{-\epsilon} + e x^{-1}\sin(2 e^{-1/2} x),\tag{7} $$ with $d\approx -0.18$, $\epsilon\approx 0.17$, $e\approx -0.097$.