The map $n \mapsto (n,0)$ from $\mathbb{Z}$ to $\mathbb{Z} \times \mathbb{Z}$ preserves addition and multiplication and does not preserve $1$, so the clause $f(1) = 1$ is not redundant

False claim: if $R$ and $S$ are rings and $f : R \to S$ satisfies $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$ for all $x, y \in R$, then $f$ is a ring homomorphism (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$); that is, clause (RH3), $f(1_R) = 1_S$, is redundant.

The map

$f : \mathbb{Z} \longrightarrow \mathbb{Z}\times\mathbb{Z}, \qquad f(n) := (n, 0)$

refutes it, $\mathbb{Z}\times\mathbb{Z}$ being the product ring (The product ring $R \times S$ with componentwise operations, its identity $(1_R, 1_S)$ and its units $R^{\times} \times S^{\times}$). It satisfies both displayed conditions and sends $1$ to $(1,0)$, which is not the identity $(1,1)$ of $\mathbb{Z}\times\mathbb{Z}$.

• The contrast with groups is the point. By (RH1) alone the map $f$ is a homomorphism of the additive groups, and for groups preservation of the identity is automatic (A group homomorphism automatically satisfies $f(e) = e'$ and $f(g^{-1}) = f(g)^{-1}$, and $f(g^{n}) = f(g)^{n}$ for every $n \in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed); indeed $f(0) = (0,0)$ here. The multiplicative structures are only monoids, and for monoids the analogous statement is false, which is exactly why Monoid homomorphism and group homomorphism imposes $f(e) = e'$ on monoid homomorphisms and Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$ imposes (RH3).

• The image is not a subring. $f(\mathbb{Z}) = \mathbb{Z}\times\{0\}$ is closed under subtraction and multiplication and does not contain $(1,1) = 1_{\mathbb{Z}\times\mathbb{Z}}$, so it fails clause (T1) of Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication — the same clause that $2\mathbb{Z}$ is closed under addition, negation and multiplication and is not a subring of $\mathbb{Z}$, because it does not contain $1$ fails, for the same reason, namely that closure alone never supplies the ambient identity. It follows that A ring homomorphism satisfies $f(0) = 0$, $f(-a) = -f(a)$ and $f(ma) = m f(a)$ for $m \in \mathbb{Z}$, carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms, whose claim 4 puts a subring as the image of a ring homomorphism, really does use (RH3).

• $(1,0)$ is an identity for the image, but not the identity of the ambient ring. For $(m,0)$ in the image, $(1,0)(m,0) = (m,0)$. So the failure is not that the image has no identity; it is that its identity is not $1_{\mathbb{Z}\times\mathbb{Z}}$, and Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication asks for the ambient identity.