The answer is no even for finitely generated groups.
Here's a construction of a finitely generated residually Hopfian, non-Hopfian group. It is even solvable (actually center-by-metabelian).

Denote by $M(u,v,x,y,z)$ the matrix $$\begin{pmatrix}u & x & z\\ 0 & v & y \\ 0 & 0 & 1\end{pmatrix}.$$

Fix a prime $p$; let $G$ be the group of matrices $M(0,t^n,x,y,z)$ for $(n,x,y,z)\in\mathbf{Z}\times\mathbf{F}_p[t,t^{-1}]^3$. Note that $G$ is finitely generated (by $\{M(0,1,0,0,0),M(0,0,1,0,0),M(0,0,0,1,0)\}$).

Let $Z$ be the central subgroup of $G$ of those matrices of the form $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t]$. Let $\alpha$ be the matrix $M(t,0,0,0,0)$ (i.e., diagonal $(t,1,1)$). Then $\alpha Z \alpha^{-1}$ is contained in $Z$ with index $p$. In particular, $\alpha$ induces a surjective, non-injective endomorphism of $G/Z$, so $G/Z$ is non-Hopfian.

**Proposition.** $G/Z$ is residually Hopfian.

I will indeed prove that it has two normal subgroup with trivial intersection, each quotient being Hopfian.

Let us "classify" quotients of $G$. Let $Z'$ be its center (the subgroup of those $M(0,0,0,0,z)$ for $z\in\mathbf{F}_p[t,t^{-1}]$.

**Claim 1:** every normal subgroup $N$ of $G$ is either contained in the center $Z'$, or contains some finite index subgroup of $Z'$.

Proof: suppose that $N$ contains an element not in $Z'$. Write $N'$ for the set of elements in $N$ of the form $M(0,0,x,y,z)$; then $N'$ is also a normal subgroup of $G$. Performing a commutator if necessary, we see that $N'$ is not contained in $G$: it contains an element of the form $M(0,0,x,y;z)$ with $(x,y)\neq (0,0)$; by symmetry we can suppose $x\neq 0$. Write $p_1,p_2,p_3$ for the projections $M(0,0,x,y;z)\mapsto x,\mapsto y,\mapsto z$. Since $N'$ is a normal subgroup, $p_1(N')$ is a nonzero $\mathbf{F}_p[t,t^{-1}]$ submodule of $\mathbf{F}_p[t,t^{-1}]$, that is, an ideal. Taking commutators with $M(0,0,0,1,0)$, we deduce that $p_3(N')$ also contains this finite index ideal. So $N$ contains some finite index subgroup of $Z'$.$\Box$

**Claim 2:** every finite-by-metabelian f.g. group $G$ is Hopfian.

Proof: every such group satisfies the max-n property (max condition on chains of normal subgroups), an old result of Ph. Hall, and max-n immediately implies Hopfian (for a surjective non-injective endomorphism, the sequence of kernels of iterates is strictly increasing). $\Box$

If $N$ is a normal subgroup of $G$ above, if $N$ contains a finite index subgroup of $Z'$, then $G/N$ is finite-by-metabelian, hence Hopfian by Claim 2.

Otherwise, $N$ is contained in $Z'$, i.e., is central. We now assume so. Since the metabelian group $G/Z'$ has trivial center [this has index 2 in a lamplighter group $C_p\wr\mathbf{Z}$, and is a lamplighter $C_p^2\wr\mathbf{Z}$], the center of $G/N$ is reduced to $Z'/N$. Let $f$ be a surjective endomorphism of $G/N$. By surjectivity, $f$ maps center into the center, and hence induces an endomorphism of the metabelian quotient $G/Z'$, which is Hopfian. Hence the inverse image of the center equals the center, and $f$ has kernel contained in the center $Z'/N$.

Up to replace $f$ with its square, we can suppose that the action of $f$ on $\mathbf{Z}$ (which appears modding out by the set of torsion elements, which is a subgroup) is trivial. Hence the action of $f$ on "$(x,y)$" (formally speaking, its action on $\mathbf{F}_p[t,t^{-1}]^2$ obtained modding out the center and restricting to the torsion subgroup) is by $\mathbf{F}_p[t,t^{-1}]$-module automorphism: it is given by some matrix $q\in\mathrm{GL}_2(\mathbf{F}_p[t,t^{-1}])$. Passing to some proper power if necessary, we can suppose that the determinant $\det(q)$ equals $t^m$ for some $m\in\mathbf{Z}$.

Shaking hands, the action of $q$ on the center $Z'$ (this doesn't make sense!) should be given by multiplication the determinant $t^m=\det(q)\in\mathbf{F}_p[t,t^{-1}]^\times$ and this should restrict the possibilities for $N$.

More precisely, for $s\in\mathbf{F}_p[t,t^{-1}]$, write $X_s=M(0,0,s,0,0)$, $Y=M(0,0,0,1,0)$, so $X_sYX_s^{-1}Y^{-1}=M(0,0,0,0,s)=Z_s$. Let $X'_s$, $Y'$ and $Z'_s$ be their images in $G/N$. So $f(Z'_s)=Z'_{F(s)}$ for some $F(s)$ which is well-defined up to addition by an element of $I=p_3(N)$.

Then computing $f$ on this commutator shows that $\det(q)s-F(s)\in I$ for every $s$. Prescribing $F$ on the canonical basis, we can suppose that $F$ is an additive homomorphism. Let $K$ be the kernel of $f$ and $J=p_3(K)$. Then $I$ is strictly contained in $J$, and $F^{-1}(I)=J$ (essentially by definition).

If $m=0$ by contradiction, we get that $F$ is identity modulo $I$, which contradicts $f(J)\subset I$.

Now suppose that $I$ is has basis $(t^n)_{n\in A}$ for some negatively unbounded subset $A\subset\mathbf{Z}$ containing $\mathbf{N}$ which in the negative is "more and more sparse" (in the sense that $A=\mathbf{N}\cup\{a_k:k\in\mathbf{N}\}$ with $a_k-a_{k-1}\to - \infty$). We combine the conditions $F(I)\subset I$ and $t^ms-F(s)\in I$, to get $t^{m+n}\in I$ for all $n\in A$, which contradicts the sparse condition.

Hence if $N=N_I$ is the group just chosen, then $G/N_I$ is Hopfian. Taking $I,I'$ sparse and with intersection reduced to $\mathbf{N}$, we get $N_I\cap N_{I'}=Z$. So $G/Z$ is residually Hopfian.