A little bit more precise: let $\mathfrak{A}$ and $\mathfrak{B}$ be two structures. Define a weak homomorphism as a function $h: \mathfrak{A} \to \mathfrak{B}$ such that the folowing conditions are satisfied:
(i) If $R_n$ is an $n$-ary relation on $A$, then, if $\langle a_1, a_2, \dots a_n \rangle \in R_n$, then there is an $n$-ary relation $S_n$ on $B$ such that $\langle h(a_1), h(a_2), \dots h(a_n) \rangle \in S_n$;
(ii) If $f_n$ is an $n$-ary function on $A$ such that that $f_n (a_1, \dots, a_n) = a$, then there's a corresponding $n$-ary function $g_n$ on $B$ such that $g_n (h(a_1), \dots, h(a_n)) = h(a)$;
(iii) If $c$ is a constant in $A$, then there's a corresponding constant $d$ in $B$ such that $h(c) = d$.
Consider now a weak homomorphism $h: \mathfrak{A} \to \mathfrak{B}$ and let $h[\mathfrak{A}]$ be the homomorphic image of $\mathfrak{A}$. Is $h[\mathfrak{A}]$ a substructure of $\mathfrak{B}$? My impulse would be to say yes, and I think I can prove it rather easily if $h$ were instead a strong homomorphism (i.e. if the converse of the implication (i) also held). As it stands, I'm a bit unsure. For instance, suppose $a_1, a_2, \dots, a_n$ are arbitrary elements of $h[\mathfrak{A}]$ in such a way that $\langle a_1, a_2, \dots, a_n \rangle \in S_n$, where $S_n$ is an arbitrary relation on $h[\mathfrak{A}]$. Then, although it's clear that each $a_i$ ($i \leq n$) is such that there's a corresponding $b_i \in A$ such that $a_i = h(b_i)$, that does not mean that there's a corresponding relation $R_n$ on $A$ such that $\langle b_1, b_2, \dots, b_n \rangle \in R_n$, for this is a weak homomorphism, not strong. And without this step, I apparently can't reach the corresponding relation on $B$. Am I missing something?
Even if we assume $\mathfrak{A}$ and $\mathfrak{B}$ have the same language, the answer is no. Consider a language with two unary function symbols $f$ and $g$. Let $\mathfrak{A}$ be a model with domain $\mathbb{N}$ where $f, g$ are each interpreted as the identity, and let $\mathfrak{B}$ be a model with domain $\mathbb{Z}$ where $f$ is interpreted as the identity but $g$ is interpeted as $x\mapsto -x$.
Then the inclusion map $\mathbb{N}\hookrightarrow\mathbb{Z}$ is a weak homomorphism from $\mathfrak{A}$ to $\mathfrak{B}$, but the image of $\mathfrak{A}$ under this map is not closed under $g$.