Given an algebraic structure $(A, f_1, f_2, \ldots)$ in the sense of universal algebra morphism are always required to respect all operations (this also includes the constants, as these are modelled as $0$-ary operations). Similar, substructures are closed under all operations, hence they include the same constants as the original algebraic structure.
A monoid has signature $(M, \cdot, 1)$, and respecting $1$ means that the identity should be preserved, similar a group homomorphism is required to preseve inversion. A submonoid is a subset that contains $1$ and so on.
As pointed out by others, we could not get rid of the requirement that the identity should be preserved, i.e. monoids does not form a full subcategory of semigroups. But as you have shown, the image of the identity forms an identity in the image of the morphism.
But sometimes we have some relations between these concepts, particular in the case of groups. Groups form a full subcategory of semigroups, for a proof see here. Similar, a finite subsets of a group is a subsemigroup iff it is a subgroup, this follows by looking at the powers of elements from the subset. So sometimes authors may define group homomorphisms in this weaker sense, or even subgroups just as subsemigroups if they are just concerned with finite groups. Something similar concerns kernels. In general these are relations defined by looking at that elements that are mapped to the same element under a homomorphims, giving rise to a congruence relation and congruence classes, again in the group case such a kernel is fully specified by just giving a single congruence class which form a normal subgroup, and almost every group theorist uses this more restrictive definition of kernel then. Also in a monoid there could be many subsemigroups that form themselve monoids or even subgroups (again look at the powers of some element), but these are not submonoids if they do not contain the identity of the whole monoid. So, sometimes we do not use all properties of the correct definitions, but in your case of monoids we do need all of them.