I succeeded in proving it myself:
Suppose we have homotopy inverses $f: X \to Y$ and $g: Y \to X$, where $X$ and $Y$ are locally connected and semi-locally simply connected (e.g. manifolds). Let $\mathcal{C}$ be a local system (a locally constant sheaf) over $Y$.
The adjoint equivalence of categories between sheaves and etale spaces restricts to an adjoint equivalence between local systems and covering spaces (this works for sheaves of sets, groups, modules etc.)
Then by homotopy invariance of fibre bundle pullbacks (on nice spaces), $f^*$ and $g^*$ will be homotopy inverses (i.e. giving an equivalence of categories) between the categories of covering spaces over $X$ and $Y$, and hence between the corresponding categories of local systems of these two spaces.
Without loss of generality, lets take $g^*$ to be the right adjoint to $f^*$ (since an equivalence of categories can be promoted to an adjoint equivalence).
Then using the natural isomorphisms $\Gamma(Y, \mathcal{C}) \cong \mathrm{Hom}(\mathbb{1}, \mathcal{C})$ and $f^{-1} \mathbf{1} \cong \mathbf{1}$ where $\mathbf{1}$ is the tensor unit (e.g. the constant sheaf of integers if we work with sheaves of abelian groups), we get natural isomorphisms
$$\Gamma(Y, \mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, \mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, g^{-1}f^{-1}\mathcal{C}) \cong \mathrm{Hom}(f^{-1}\mathbf{1}, f^{-1}\mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, f^{-1}\mathcal{C}) \cong \Gamma(X, f^{-1}\mathcal{C}).$$
The (homology of the) right derived functor $R\Gamma(X, -)$ is sheaf cohomology, so we are done.
There's no need to compute a full resolution of $\Bbb Z$ - a short exact sequence is enough to solve the problem if you're clever about it.
Let $\def\cF{\mathcal{F}} \def\ZZ{\mathbb{Z}} \def\RR{\mathbb{R}} \def\cQ{\mathcal{Q}} \def\G{\Gamma} \def\coker{\operatorname{coker}} \cF$ be the flasque sheaf which assigns to each $U\subset S^1$ the set of all functions $U\to\RR$.
Embed $\ZZ\to\cF$ in the obvious way and let $\cQ$ be the quotient.
Taking the long exact sequence in cohomology arising from $$0\to \ZZ\to \cF\to \cQ\to 0,$$ we note that $H^1(S^1,\cF)=0$ implying $H^1(S^1,\ZZ)=\coker(\G(S^1,\cF)\to\G(S^1,\cQ))$.
By exercise II.1.3, any section $s\in\G(S^1,\cQ)$ is the image of a family $\{(s_i,U_i)\}_{i\in I}$ with $s_i\in\cF(U_i)$ where $U_i$ form an open cover of $S^1$ and $(s_i-s_j)|_{U_i\cap U_j}$ is a section of $\ZZ_{U_i\cap U_j}$.
Since $S^1$ is compact, we may assume $I$ is finite; after subdividing, throwing away redundant elements, and reordering we may assume that our cover consists of connected open subsets so that $U_i$ only intersects $U_{i-1}$ and $U_{i+1}$ with indices interpreted modulo $|I|$.
Now I claim that it suffices to consider $|I|=3$.
Let $n_{i+1}$ be the value of $s_{i+1}-s_i$ on $U_i\cap U_{i+1}$.
Replacing $s_{i+1}$ with $s_{i+1}-n_{i+1}$, which does not change the image of $s_{i+1}$ in $\cQ(U_{i+1})$, we see that $s_i=s_{i+1}$ on $U_i\cap U_{i+1}$.
Therefore we can glue $s_i$ and $s_{i+1}$ to form a section of $\cF$ over $U_i\cup U_{i+1}$ without changing its image in $\cQ$.
Repeating this process for $i=3,\cdots,|I|-1$, we see that we can glue the sections $s_i$ on $U_3\cup\cdots\cup U_{|I|}$ so that we're only looking at $\{(s_1,U_1),(s_2,U_2),(s_3,U_3\cup\cdots\cup U_{|I|})\}$.
If we have a section $\{(s_1,U_1),(s_2,U_2),(s_3,U_3)\}$, by the same logic we may assume that $s_1=s_2$ on $U_1\cap U_2$ and $s_2=s_3$ on $U_2\cap U_3$.
Therefore up to adding a global section of $\cF$, the global sections of $\cQ$ are exactly those of the form $\{(0,U_1),(0,U_2),(n,U_3)\}$ for $n\in\ZZ$ and opens $U_i$ satisfying our ordering and intersection assumptions.
Since any two such sections are equivalent up to an element of $\G(S^1,\cF)$ iff their $n$s match, we see that the cokernel is exactly $\ZZ$.
Best Answer
Call a sheaf on $\mathbb{R}$ interval-flasque if for every inclusion of open intervals $(a, b) \subseteq (a', b')$, the restriction map $\Gamma((a', b'), \mathscr{F}) \to \Gamma((a, b), \mathscr{F})$ is surjective. (Here, we allow $a = -\infty$ and/or $b = \infty$ and similarly for $a', b'$.) We then claim that any interval-flasque sheaf of abelian groups on $\mathbb{R}$ is acyclic in $\mathfrak{Ab}(\mathbb{R})$. Since in particular, the constant sheaf $\mathbb{Z}$ on $\mathbb{R}$ is clearly interval-flasque, that will imply the desired result.
The proof proceeds similarly to the proof for generally flasque sheaves. Here I'll follow the outline of the facts that the proof from Hartshorne's Algebraic Geometry uses:
If $0 \to \mathscr{F}' \to \mathscr{F} \to \mathscr{F}'' \to 0$ is a short exact sequence, and $\mathscr{F}'$ is interval-flasque, then $0 \to \Gamma(U, \mathscr{F}') \to \Gamma(U, \mathscr{F}) \to \Gamma(U, \mathscr{F}'') \to 0$ is exact for every open interval $U$. Here, in using Zorn's Lemma to show there exists a maximal preimage of $x \in \Gamma(U, \mathscr{F}'')$, you will need to use the fact that the union of a chain of intervals is again an interval. Then, in showing that a preimage on a strict subinterval can be extended (and therefore, a maximal preimage must be a section over the full larger interval), you will need to use the fact that a nonempty intersection of two intervals is an interval, and likewise for the union of two overlapping intervals.
If $0 \to \mathscr{F}' \to \mathscr{F} \to \mathscr{F}'' \to 0$ is a short exact sequence, and $\mathscr{F}'$ and $\mathscr{F}$ are both interval-flasque, then so is $\mathscr{F}''$. Here, the proof is pretty much the same as for the generally flasque case: use the previous fact, along with the Snake Lemma on the obvious morphism between short exact sequences of abelian groups.
Any injective object of $\mathfrak{Ab}(\mathbb{R})$ is interval-flasque. Well, we know it's flasque, so in particular it's interval-flasque.
So now, the standard proof that flasque sheaves are acyclic will also work for interval-flasque sheaves, with just a global replacement of "flasque" with "interval-flasque" in the proof. Note that we do need the fact here that $\mathbb{R} = (-\infty, \infty)$ is considered an open interval in the preceding facts.
Now, it might be interesting to see where this proof fails if you try to apply it on $S^1$ instead of $\mathbb{R}$. In particular, what goes wrong if you try to replace "open intervals in $\mathbb{R}$" with "open arcs in $S^1$"? What goes wrong if you try to fix this failure by restricting to open arcs of length less than $\pi$?