Yes. [The tameness of $f$ makes our life much easier.]
It will be convenient for the proof to assume that $C$ is pure: that is, every vertex and edge of $C$ lies in some triangle of $C$. If this is not the case, then glue triangles on until it is the case.
Take $D = f(C)$; since $f$ is piecewise linear $D$ is tame and we can work with $D$ instead of $C$. Let $N$ be a very small regular neighbourhood of $D$. Note that $D$ is a spine for $N$ in the sense that $N - D \cong S \times (0, 1]$ (PL homeo). Let $\tau_D : N \to D$ be the resulting deformation retraction. (In fact, we can here get away with using nearest point projection.) So $N \subset S^3$ is a PL sub-three-manifold and $S = \partial N$ is a separating surface. Note that $S$ is need not be connected and may have two-sphere components.
Define $M$ to be the closure of $S^3 - N$. Note that $S = \partial M$.
Let $E$ be a special spine for $M$. Again, we will assume that $E$ is pure. (Note that this means that we will have use Bing's house with two rooms, or similar, as special spines for three-ball components of $M$.) Thus $M - E \cong S \times (0, 1]$. Again let $\tau_E : M \to E$ be the resulting deformation retraction. We can no longer use closest point projection, but we still arrange that $\tau_E$ is PL.
Let $\Gamma$ be a spine for the surface $S$; that is, $S - \Gamma$ is a collection of disks. Again we assume that $\Gamma$ is pure - so while $\Gamma$ may have vertices of degree one, it does not have isolated vertices. Also, we isotope $\Gamma$ slightly so that it is transverse to $(\tau_D|S)^{-1}(D^{(1)})$ and also to $(\tau_E|S)^{-1}(E^{(1)})$.
We build a two-complex from $$D \,\, \sqcup \,\, \Gamma \times [-1, 1] \,\, \sqcup \,\, E$$ by attaching $\Gamma \times \{-1\}$ to $D$ via $\tau_D$ and attaching $\Gamma \times \{1\}$ to $E$ via $\tau_E$. This is the desired complex $C'$; the map $f'$ is inclusion. Note that $S^3 - C'$ is a collection of (open) three-balls. That is, $C'$ is a spine for $S^3$. Thus $C'$ is simply connected.
