It is easy to see that the dual space of a bornological space $V$ (i.e. the space of bounded linear functionals) may be zero (just take any vector space with the power set as bornology). Hence in general, you cannot conclude that $v=0$ from the statement that $\langle\xi, v\rangle = 0$ for all $\xi \in V^\prime$: The dual space may not have "enough" linear functionals
The question is: Can we conclude $z=0$ for $z \in V\hat{\otimes}V^\prime$ from the statement that $\langle A, z \rangle = 0$?, where $\hat{\otimes}$ denotes the completed bornological tensor product. That is, does at least the space $V \hat{\otimes} V^\prime$ have enough linear functionals in this sense?
Remark: Note that the dual of the complete tensor product $V \hat{\otimes} V^\prime$ is given by $$ (V \hat{\otimes} V^\prime)^\prime \cong L(V,V^{\prime\prime}).$$ This makes the dual pretty accessible. I am not sure how to use this though.
