I've deleted this answer, since it completely doesn't work. If I can fix it, I'll replace it.

The answer is no, if I understand correctly. (I use $E$ instead of $\in$, here, since the latter is fairly meaning-loaded to me.)

Consider the set of sentences $\Gamma=\lbrace \forall x\exists y(xEy), \forall x(\neg xEx) \rbrace$. Then if $V$ has two elements and $W$ has one element, and $f$ is the unique function from $V$ to $W$, clearly $\Gamma$ is consistent but $f(\Gamma)$ is inconsistent.

Does this answer your question?

Of course, the key point above is that we have negation. Without negation, the answer to your question is "yes:"

Let me rephrase your question in more semantic terms, which I find easier to deal with. If I understand correctly, you are considering the signature $\Sigma=\lbrace E\rbrace$, where $E$ is a binary relation; then, given sets $V$, $W$ and a map (not a homomorphism) $f: V\rightarrow W$, you are asking whether, given a $\Sigma$-structure $\mathcal{M}$ with domain $V$ and a consistent set $\Gamma$ of sentences with parameters from $V$ which do not use the symbol "=" and are true in $\mathcal{M}$, we can find some $\Sigma$-structure $\mathcal{N}$ with domain $W$ such that $$ \forall \varphi(\overline{a})\in \Gamma, \mathcal{N}\models \varphi(f(\overline{a})).$$ (Quantifying over sets of sentences holding in $\Sigma$-structures should be the same as quantifying over consistent sets of sentences, as long as things aren't terrible.) Without loss of generality, $\Gamma$ is just the set of all sentences with parameters from $V$ true in $\mathcal{M}$; that is, we can just quantify over equality- and negation-free diagrams.

Let $A=ran(f)$, and let $\mathcal{A}=f(\mathcal{M})$ be the structure on $W$ gotten by moving over the structure $\mathcal{M}$, so that $f$ (with new codomain) is a homomorphism from $\mathcal{M}$ to $\mathcal{A}$. $$\text{(Formally, we set $\mathcal{A}=(A; \lbrace (x, y): \mathcal{M}\models f^{-1}(x)Ef^{-1}(y)\rbrace)$.)}$$ $\mathcal{A}$ satisfies $D_{enf}(\mathcal{M})$, the equality- and negation-free diagram of $\mathcal{M}$, by an easy induction.

Clearly if $A=W$, we're done. So suppose $A\not=W$. Now fix $c\in \mathcal{A}$, let $B=W-A$, and consider the structure $\mathcal{N}$ on $W$ gotten from $\mathcal{A}$ by creating several identical copies of $c$. Formally, for $b\in B, x\in W$, we make $bEx$ hold iff $\mathcal{A}\models cEx$, and we make $xEb$ hold iff $\mathcal{A}\models xEc$.

$\mathcal{N}$ is clearly a $\Sigma$-structure with domain $W$; and by induction on the complexity of $\varphi(\overline{a})\in D_{enf}(\mathcal{M})$, we have $\mathcal{N}\models\varphi(f(\overline{a}))$. So we're done.