Here is some partial progress which I decided to hive off from the question itself. We use the series definition of the generalized hypergeometric function to write the first term as
$${_2}F_2 \left(\begin{array}{c}1/2,-\epsilon/2\\ 3/2,-\epsilon \end{array}\Bigg{\vert} -1\right)=\sum_{k=0}^\infty \frac{ (1/2)_k} {(3/2)_k}\frac{(-\epsilon/2)_k}{(-\epsilon)_k}\frac{(-1)^k}{k!}$$ where $(a)_k$ is the Pochhammer symbol. The first ratio can be simplified to $1/(2k+1)$ without issue, but the second ratio is problematic. If we naively take $\epsilon=0$, then Mathematica compute it to be identically equal to $1$. If we instead first fix some particular integer $k$, Mathematica now concludes that the ratio converges to $1/2$ as $\epsilon\to 0$. Yet even this is likely too naive: If I hold $\epsilon>0$ fixed and take $k\to\infty$, then the ratio diverges. Similarly, the second term can be written as
\begin{align}
\frac16 {_2}F_2 \left(\begin{array}{c}3/2,1+\epsilon/2\\ 5/2,2+\epsilon \end{array}\Bigg{\vert} -1\right)
&=\frac16 \sum_{k=0}^\infty \frac{ (3/2)_k} {(5/2)_k}\frac{(1+\epsilon/2)_k}{(2+\epsilon)_k}\frac{(-1)^k}{k!}\\
&=\sum_{k=0}^\infty \frac{ 1} {6+4k}\frac{(1+\epsilon/2)_k}{(2+\epsilon)_k}\frac{(-1)^k}{k!}
\end{align}
and thus the second ratio is again problematic. The issue thus likely seems to be that of order of limits: one should hold $\epsilon\neq 0$ fixed, sum $k=1$ to $\infty$, and only then take $\epsilon\to 0$. (But this is not the same as a proof!)
Finally, let
$$S_n(\epsilon)=\sum_{k=0}^{n-1}\left[\frac{1} {2k+1}\frac{(-\epsilon/2)_k}{(-\epsilon)_k}+\frac{ 1} {6+4k}\frac{(1+\epsilon/2)_k}{(2+\epsilon)_k}\right]\frac{(-1)^k}{k!},$$ i.e., the $n$th combined partial sum of the two hypergeometric series. Based on Mathematica, I conjecture that
$$\lim_{\epsilon\to 0} S_n(\epsilon)=1-\frac12 \frac{(-1)^n}{(2n+1)n!}$$
which would seem enough to establish $f(0)=1$. (I use weasel words here because I'm paranoid about order of limits.) Curiously, the sequence $\{(2n+1)n!\}_{n=0}^\infty$ is already extent on OEIS as A007680. The principal reference is its role in the series expansion of the error function:
$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)k!}$$ Given the earlier appearance of the error function, it seems unlikely that this could be coincidental. But I can't immediately see how the two appearances are related. (Also recall the well-known fact that, while the above series for the error function does converge, it does so quite slowly: One has to go out until $k>x^2-2$ before the series satisfies the conditions of the alternating series test. This seems of a kind with the sensitivity to limit ordering displayed thus far.)
I think I've resolved the Mathematica issue to my satisfaction. First, playing around with Mathematica further the limit $\epsilon\to 0$ is not actually problematic for the second hypergeometric function. Hence we use Mathematica to obtain to
$$\sum_{k=0}^\infty \frac{ 1} {6+4k}\frac{(1)_k}{(2)_k}\frac{(-1)^k}{k!}=\frac12-\frac{\sqrt{\pi}}{4}\operatorname{erf}(1) $$
This leaves the first term. If we take $\epsilon\to 0$ immediately, the ratio of Pochammer symbols evaluates to $1$ and the hypergeometric series becomes
$$\sum_{k=0}^\infty \frac{1} {2k+1}\frac{(-1)^k}{k!}=\frac{\sqrt{\pi}}{2}\operatorname{erf}(1)$$
If we first take $k=0$, then $(a)_0=1$ and the ratio is again 1. But for any other integer, we have
$$\frac{(-\epsilon/2)_k}{(-\epsilon)_k}=\frac{-\epsilon/2}{-\epsilon}\frac{1-\epsilon/2}{1-\epsilon}\cdots \frac{n-1-\epsilon/2}{n-1-\epsilon}$$
In the limit $\epsilon\to 0$, every factor cancels aside from the first. Hence $\frac{(-\epsilon/2)_k}{(-\epsilon)_k}\to 1/2$ if $k$ is a positive integer. Therefore, the correct summation is
$$1+\sum_{k=1}^\infty \frac{1} {2(2k+1)}\frac{(-1)^k}{k!}=\frac12 + \frac{\sqrt{\pi}}{4}\operatorname{erf}(1)$$
This error function is then able to cancel that from the second term, leaving a total of $1/2+1/2$. Expressed as a single summation, we have
$$\sum_{k=0}^\infty\left[\frac{1}{2k+1}+\frac{1}{(2k+3)(k+1)}\right]\frac{(-1)^k}{k!}=1$$
which Mathematica validates. Indeed, we seem to have more generally
$$\sum_{k=0}^{n-1}\left[\frac{1}{2k+1}+\frac{1}{(2k+3)(k+1)}\right]\frac{(-1)^k}{k!}=1-\frac{1}{(2n+1)n!}$$
for the $n$th partial sum. This form suggests the series can be made telescopic but I don't immediately see it.
