Proposition. There is no minimiser.
Proof: An equivalent version of your problem is as follows.
Statement. We wish to minimise $$\int_{-\infty}^\infty xe^{q(x)-x^2}\,dx$$ where $q(x)$ is a strictly increasing function with $q(a)=0$ for a given $a\in\Bbb R$ and $$\lim_{x\to\pm\infty}q(x)=\pm\infty,\quad\lim_{x\to+\infty}q'(x)e^{q(x)-x}=+\infty.$$
The proof for the equivalence is provided at the end of this post.
Now, since $\int_{-\infty}^\infty xe^{q(x)-x^2}\,dx$ is positive by the original integral $\int_0^\infty g'(x)f(x)\,dx>0$, it suffices to choose a sequence of functions $q_A(x)$ such that the integral becomes arbitrarily close to zero.
This is shown by Iosif Pinelis on MathOverflow where we take $$q_A(x):=k(x-a)\,1_{x\le a+A}+(kA+2(x-A))\,1_{x>a+A}$$ for some $k,A\in(0,\infty)$. By the dominated convergence theorem, we have $$\lim_{k\to0^+}\lim_{A\to+\infty}\int_{-\infty}^\infty xe^{q_A(x)-x^2}\,dx=\lim_{k\to0^+}\int_{-\infty}^\infty xe^{k(x-a)-x^2}\,dx=\int_{-\infty}^\infty xe^{-x^2}=0$$ and so we have a zero infimum. $\quad\square$
Proof of Statement: Let $g=e^h$ so that $g'=h'e^h$. Then the constraints become $$h'(x)>0,h(1)=0,\lim_{x\to0}h(x)=-\infty,\lim_{x\to\infty}h(x)=+\infty,\lim_{x\to\infty}h'(x)e^{h(x)}=+\infty$$ and the functional is \begin{align}\int_0^\infty g'f\,dx&=[gf]_0^\infty-\int_0^\infty gf'\,dx\\&=\frac1{\sigma\sqrt{2\pi}}\left(\lim_{x\to\infty}e^{h(x)-\log x-\frac{(\log x-\mu)^2}{2\sigma^2}}-\int_0^\infty\frac{e^{h(x)-\frac{(\log x-\mu)^2}{2\sigma^2}}(\mu-\sigma^2-\log x)}{\sigma^2 x^2}\,dx\right)\end{align} where we assume the limit exists. Let $h(x)=p(u)$ where $x=e^u$. Then $$p'(u)>0,p(0)=0,\lim_{u\to\pm\infty}p(u)=\pm\infty,\lim_{u\to+\infty}p'(u)e^{p(u)-u}=+\infty$$ and \begin{align}\sigma\sqrt{2\pi}\int_0^\infty g'f\,dx&=\lim_{u\to\infty}e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}+\frac1{\sigma^2}\int_{-\infty}^\infty e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}(u+\sigma^2-\mu)\,du.\end{align} For both $\int_0^\infty$ and $\int_{-\infty}^0$ to be finite, we must have $$\lim_{u\to\pm\infty}p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}=-\infty\implies p(u)=o(u^2)$$ since $p'(u)>0$ for all $u$ means $p$ cannot have a quadratic term. So \begin{align}\sigma^3\sqrt{2\pi}\int_0^\infty g'f\,dx&=\int_{-\infty}^\infty e^{p(u)-u-\frac{(u-\mu)^2}{2\sigma^2}}(u+\sigma^2-\mu)\,du\\&=\frac{e^{\sigma^2/2-\mu}}{2\pi\sigma^2}\int_{-\infty}^\infty ve^{q(v)-v^2}\,dv\end{align} where $q(u)=p(u/\sqrt{2\sigma^2}-\sigma^2+\mu)$ under the same constraints as $p(u)$ except $q((\sigma^2-\mu)\sqrt{2\sigma^2})=0$. In the statement, we simply define $a:=(\sigma^2-\mu)\sqrt{2\sigma^2}$. $\quad\square$
