The axioms are the Field axioms. As a hint, consider the analogous problem of showing that the Field axioms do not imply that $\:1 + 1 \ne 0.\:$ If the implication held true then $1 + 1 \ne 0\:$ would be true in every field. To refute the implication it suffices to find a counterexample: a field with $1 + 1= 0.$ But everyone knows such a field, namely Parity Arithmetic, i.e. the integers modulo $2$, which one easily checks satisfies all the Field axioms, using the operation tables below, e.g. the commutative laws hold because the operation tables are symmetric.
$\begin{array}{rcl} \rm Parity\ Arithmetic & &\rm\!\!\! modulo\ 2 \\ \begin{array}{|c|c|c|}\hline + & \bf\text{even} &\rm\bf odd\\\hline \bf\text{even} & \text{even} & \text{odd}\\\hline \rm\bf odd & \text{odd} &\rm even\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline + & \bf 0 & \bf 1\\\hline \bf 0 & 0 & 1\\\hline \bf 1 & 1 & 0\\\hline \end{array} \\ \begin{array}{|c|c|c|}\hline \times & \bf\text{even} &\rm odd\\\hline \bf\text{even} & \text{even} & \text{even}\\\hline \rm\bf odd & \text{even} &\rm odd\\\hline \end{array} \!\!\!\!\!\!\!& \iff &\!\!\!\!\!\!\! \begin{array}{|c|c|c|}\hline \times & \bf 0 & \bf 1 \\\hline \bf 0 & 0 & 0\\\hline \bf 1 & 0 & 1\\\hline \end{array} \end{array} $
Your problem is simpler since a counterexample has $\:1 = 0,\:$ hence $\rm\: x = 1\cdot x = 0\cdot x = 0,\:$ i.e. every element $= 0,\:$ so the structure has only the single element $0.\:$ The axiom for inverses is vacuously satisfied, since there are no nonzero elements to check for inverses. All other axioms are universal, i.e. identities such as $\rm\:\forall\, x,y\!:\ x+y = y + x,\:$ and they hold true for all elements simply because there is only a single value $\:0\:$ for the operations to take, so it can only evaluate to $\:0 = 0\:$ (vs. possibly $\rm\:1 = 0\:$ when verifying the two-element field above), see the operation tables below.
$\rm\begin{array}{} Null\ Arithmetic\ (mod\ 1)\\ \begin{array}{|c|c|}\hline + & \bf 0 \\\hline \bf 0 & 0 \\\hline \end{array}\quad \begin{array}{|c|c|}\hline * & \bf 0 \\\hline \bf 0 & 0 \\\hline \end{array} \end{array} $