EDIT: Now it has a chance of making sense.

I think the equivalence

"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to a non-unit $\Longleftrightarrow$ $a$ is nilpotent"

cannot be true (though $\Longleftarrow$ holds, of course). Otherwise, it would yield that whenever $a$ is nilpotent, then so is $ua$ for any unit $u$ of $A$, but this is not satisfied in the ring $\mathbb{Z}\left\langle X,Y,Z\right\rangle / \left(X^2,YZ-1,ZY-1\right)$ (the ideal is two-sided). The counterexample is $u=Y$, $a=X$, $v=1$ (while $a=X$ is nilpotent, $ua=YX$ isn't).

EDIT: Now it has a chance of making sense.

I think the equivalence

"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to a non-unit $\Longleftrightarrow$ $a$ is nilpotent"

cannot be true (though $\Longleftarrow$ holds, of course). Otherwise, it would yield that whenever $a$ is nilpotent, then so is $ua$ for any unit $u$ of $A$, but this is not satisfied in the ring $\mathbb{Z}\left\langle X,Y,Z\right\rangle / \left(X^2,YZ-1,ZY-1\right)$ (the ideal is two-sided). The counterexample is $u=Y$, $a=X$, $v=1$ (while $a=X$ is nilpotent, $ua=YX$ isn't).

The same counterexample proves that

"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to an element neither left-invertible nor right-invertible $\Longleftrightarrow$ $a$ is nilpotent"

must be wrong.

I think the equivalence

"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to a non-unit $\Longleftrightarrow$ $a$ is nilpotent"

cannot be true (though $\Longleftarrow$ holds, of course). Otherwise, it would yield that whenever $a$ is nilpotent, then so is $uav$ for any $u$ and $v$ from $A$, but this is not satisfied in the ring $\mathbb{Z}\left\langle X,Y\right\rangle / \left(X^2\right)$ (the ideal is two-sided). The counterexample is $u=Y$, $a=X$, $v=1$ (while $a=X$ is nilpotent, $uav=YX$ isn't).

EDIT: Now it has a chance of making sense.

I think the equivalence

"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to a non-unit $\Longleftrightarrow$ $a$ is nilpotent"

cannot be true (though $\Longleftarrow$ holds, of course). Otherwise, it would yield that whenever $a$ is nilpotent, then so is $ua$ for any unit $u$ of $A$, but this is not satisfied in the ring $\mathbb{Z}\left\langle X,Y,Z\right\rangle / \left(X^2,YZ-1,ZY-1\right)$ (the ideal is two-sided). The counterexample is $u=Y$, $a=X$, $v=1$ (while $a=X$ is nilpotent, $ua=YX$ isn't).