We have to distinguish between two algebraic structures with appropriate homomorphisms between them (in fact, they constitute categories)
$\bullet$ Rings (which I always assume to be unital) with homomorphisms of rings.
As always, a homomorphism of blupp preserves the whole structure of blupp, so in particular for blupp=Ring the unit is preserved by definition. This definition is (or should be ...) universally accepted.
$\bullet$ Rngs or non-unital rings with homomorphisms of rngs
Following the general principle, a homomorphism of rngs preserves the addition and the multiplication (and then also additive inverses and the zero), but not the unit because there is none.
There is a so-called forgetful functor from Rings to Rngs which forgets the unit. Notice that if $R$ is a ring, the underlying rng $|R|$ is a different object (and similarly for homomorphisms). This is usually ignored in textbooks and lectures, leading to utter confusions. Similar problems arise with other forgetful functors. These problems will end immediately when we take forgetful functors seriously. Unfortunately, some authors consider rings with unit but consider also non-unital homomorphisms between them, which doesn't make sense at all because this actually ignores the general principles of universal algebra, secretly applies the forgetful functor all the time, and doesn't take rings seriously. And some authors even don't consider $0$ as a ring because "it has no unit", sic! Of course the zero ring is a ring. It's quite cold these days, so books following this approach would make a warm fire ...
Anyway, now the question is easily answered:
Let $f : R \to R'$ be a homomorphism of rings (thus preserving the unit by definition). Then $f=0$ iff $R'=0$. It doesn't matter if $R$ is a field or not, the condition doesn't depend on $R$ at all.
On the other hand, there is always a zero homomorphism of the underlying rngs $0 : |R| \to |R'|$. Notice, again, that these are different objects than the rings themselves!
Best Answer
We will denote by $\overline{f}$ the image of $f\in\mathbb{R}[x]$ under the canonical projection onto the quotient $A=\mathbb{R}[x]/\langle (x+1)x\rangle$.
Indeed, $A$ has nontrivial zero divisors e.g. $\overline{x+1}$ and $\overline{x}$.
Furthermore, $A$ has no nontrivial nilpotent elements. Indeed, let $\overline{f}$ be a nontrivial nilpotent element. Then $f^n$ is divisible by $(x+1)x$ for some positive integer $n$. Since $x+1$ and $x$ are both prime in $\mathbb{R}[x]$, and they have greatest common divisor $1$, $f$ must be divisible by $(x+1)x$.
Another way to see this is via the following isomorphisms. $$A\cong \mathbb{R}[x]/\langle x+ 1\rangle\oplus\mathbb{R}[x]/\langle x\rangle\cong \mathbb{R}\oplus\mathbb{R}.$$ The first isomorphism is given by the Chinese remainder theorem, and the second isomorphism is induced by evaluation homomorphisms $\mathbb{R}[x]\to\mathbb{R}$ (at $-1$ for the first component and $0$ for the second).