Does there exist unique $u \in V$ satisfying integral equation?

PARTIAL ANSWER:

The main difficulty here is (IMHO) to prove that all integrals make sense. Once this is accomplished, the problem SHOULD be standard and solvable, e.g., via Lax-Milgram's theorem.

Now in the case of $$\int_0^1 \frac{f(x)v(x)}{x^2}\, dx$$ I would proceed as follows. The task is to show that $\frac{v(x)}{x}\in L^2$. Use the fact that
$$v(x)=\int_0^x v'(y)\, dy.$$ So $$ \begin{split} \int_0^1 v(x)\frac1x\, dx &= \int_0^1 \left(\int_0^x v'(y)\, dy\right)\, d\log x \\ & = \int_0^1 -v'(x)\log x\, dx +\log 1 \int_0^1 v'(y)\, dy - \lim_{x\to 0} \log x \int_0^x v'(y)\, dy. \end{split} $$ By l'Hôpital's rule the last limit is $0$. So the integral is finite and this shows that $\frac{v(x)}{x}\in L^2$.

The same reasoning shows that the term $\frac{u(x)}x\in L^2$ too. We are therefore dealing with the weak formulation of the Sturm-Liouville problem $$ \begin{cases} u'' +\frac{1}{x^2}u' = F, & F\overset{\text{def}}{=}\frac{f}{x^2} \\ u(0)=0 \\ u'(1)=0 \end{cases} $$

Now I should check if Lax-Milgram's theorem is directly applicable or if we need some more ideas. The fact that $F$ is not necessarily in $L^2$ makes me tend towards the second scenario.

But I have ran out of time, so I am just writing down this community wiki hoping that it can be of some help.

If your weak equation holds, then it holds for all $v\in C^{\infty}_c(0,1)$ (compactly supported,) which implies that $u'$ must be absolutely continuous and satisfy $$ -u''+\frac{1}{x^2}u = \frac{1}{x^2}f. $$ The solutions of $-u''+\frac{1}{x^2}u=0$ have the form $x^{p}$ where $p(p-1)-1=0$. So, $$ (p-1/2)^2-5/4=0,\\ p = \frac{1}{2}\pm\frac{\sqrt{5}}{2},\\ p_1 = \frac{1+\sqrt{5}}{2},\;\;p_2=\frac{1-\sqrt{5}}{2}. $$ The Wronskian of the solutions $x^{p_1}$, $x^{p_2}$ is $$ W(x^{p_1},x^{p_2})=x^{p_1+p_2-1}(p_1-p_2)=\sqrt{5}. $$ Consider the function $g$ given by $$ -\sqrt{5}g(x) = x^{p_2}\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt+x^{p_1}\int_{x}^{1}t^{p_2}\frac{f(t)}{t^2}dt. $$ This function is defined on $(0,1)$ because $$ t^{p_1}\frac{f(t)}{t^2}=t^{p_1-1}\frac{f(t)}{t}\in L^1. $$ $g$ is a Green function solution with \begin{align} -\sqrt{5}g'(x) & = (x^{p_2})'\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt +(x^{p_1})'\int_{x}^{1}(t^{p_2}-t^{p_1})\frac{f(t)}{t^2}dt \\ -\sqrt{5}g''(x) & = (x^{p_2})''\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt +(t^{p_1})''\int_{x}^{1}t^{p_1}\frac{f(t)}{t^2}dt \\ & + W(x^{p_1},x^{p_2})\frac{f(x)}{x^2} \end{align} Therefore, $$ \sqrt{5}\left(-g''+\frac{1}{x^2}g\right) = \sqrt{5}\frac{f(x)}{x^2}. $$ Take a look at the behavior of $g$ near $x=0$. The first integral term for $\sqrt{5}g$ is bounded using Cauchy-Schwarz by $$ x^{p_2}\left(\int_{0}^{x}t^{2p_1-2}\right)^{1/2}\left(\int_{0}^{x}\left(\frac{f(t)}{t}\right)^{2}dt\right)^{1/2} \\ \le x^{p_2}\frac{x^{p_1-1/2}}{\sqrt{2p_1-1}}k(x) \\ = \frac{\sqrt{x}}{\sqrt{2p_1-1}}k(x), $$ where $k(x)=\|\chi_{[0,x]}(t)f(t)/t\|\rightarrow 0$ as $x\rightarrow 0$. The second integral term for $\sqrt{5}g$ is bounded by $$ x^{p_1}\left(\int_{x}^{1}(t^{p_2-1})^2dt\right)^{1/2}\|f(t)/t\| \\ \le C x^{p_1}x^{p_2-1/2}=C\sqrt{x}. $$ Therefore $g$ vanishes at $x=0$.

Any other twice absolutely continuous solution of $-u''+\frac{1}{x^{2}}u=\frac{1}{x^2}f$ on $(0,1)$ must have the form $$ u = g + Ax^{p_1}+Bx^{p_2}, $$ However $B=0$ must hold in order for $u$ to vanish at $x=0$. By choosing $B$ appropriately, you can arrange for $u'(1)=0$. Then, when you integrate your weak equation by parts, you get $$ \int_{0}^{1}(-u'')vdx = \left.\int_{0}^{1}u'v'dx+u'v\right|_{0}^{1} = \int_{0}^{1}u'v'dx,\;\;\; v\in V. $$ (You will need to estimate $u'$ near $0$ and use the properties of $v\in V$, but the expression is given above, and the techniques are the same.) So this $u$ is a weak solution. And you can see that it is unique by the classical construction given above. This may not be what you hoped for, but I think it's a full solution. :) The answer is: Yes, there is a unique solution as posed.