I assume that the reason you want a non-inductive proof is that induction requires you to know your goal before you start, but there's actually a simple way to find out what the result could be. Assume that the general identity is of the form
$$F(-n,\vec{b};\vec{c};1-z) = \alpha_n(\vec{b},\vec{c}) F(-n,\vec{u};\vec{v};z)$$ where $\alpha_n$ is independent of $z$. Then we have an identity of polynomials, which is just an identity of coefficients. Expanding
$$\sum_{k=0}^n \frac{(-n)_k \prod_{b_i \in \vec{b}}(b_i)_k}{\prod_{c_i \in \vec{c}}(c_i)_k} \frac{(1-z)^k}{k!} = \alpha_n(\vec{b},\vec{c}) \sum_{j=0}^n \frac{(-n)_j \prod_{u_i \in \vec{u}}(u_i)_j}{\prod_{v_i \in \vec{v}}(v_i)_j} \frac{z^j}{j!}$$
we get that for $0 \le j \le n$,
$$(-1)^j \sum_{k=j}^n \frac{(-n+j)_{k-j} \prod_{b_i \in \vec{b}}(b_i)_k}{(k-j)! \prod_{c_i \in \vec{c}}(c_i)_k} = \alpha_n(\vec{b},\vec{c}) \frac{\prod_{u_i \in \vec{u}}(u_i)_j}{\prod_{v_i \in \vec{v}}(v_i)_j}$$
From the case $j=n$ we derive
$$\alpha_n(\vec{b},\vec{c}) = (-1)^n \frac{\prod_{b_i \in \vec{b}}(b_i)_n}{\prod_{c_i \in \vec{c}}(c_i)_n} \frac{\prod_{v_i \in \vec{v}}(v_i)_n}{\prod_{u_i \in \vec{u}}(u_i)_n}$$
From the case $j=n-1$ we derive
$$(-1)^{n-1} \frac{\prod_{b_i \in \vec{b}}(b_i)_{n-1}}{\prod_{c_i \in \vec{c}}(c_i)_{n-1}} \left(1 - \frac{\prod_{b_i \in \vec{b}} (b_i+n-1)}{\prod_{c_i \in \vec{c}} (c_i+n-1)} \right) = \alpha_n(\vec{b},\vec{c}) \frac{\prod_{u_i \in \vec{u}}(u_i)_{n-1}}{\prod_{v_i \in \vec{v}}(v_i)_{n-1}}$$
Putting them together,
$$\frac{\prod_{v_i \in \vec{v}} (v_i+n-1)}{\prod_{u_i \in \vec{u}} (u_i+n-1)} = 1 - \frac{\prod_{c_i \in \vec{c}} (c_i+n-1)}{\prod_{b_i \in \vec{b}} (b_i+n-1)}$$
In the case that $\vec{b} = b$ and $\vec{c} = c$, the RHS is simply $1 - \frac{c+n-1}{b+n-1} = \frac{b-c}{b+n-1}$ and identifying $v+n-1 = b-c$, $u+n-1=b+n-1$ gives the quoted theorem. If we have e.g. $\vec{b} = (b_1,b_2)$, $\vec{c} = (c_1,c_2)$ then the RHS is $$\frac{(b_1+n-1)(b_2+n-1)-(c_1+n-1)(c_2+n-1)}{(b_1+n-1)(b_2+n-1)} = \frac{(b_1 + b_2 - c_1 - c_2)n + (b_1-1)(b_2-1) - (c_1-1)(c_2-1)}{(b_1+n-1)(b_2+n-1)}$$
which is suggestive that the most general identity possible requires $b_1 + b_2 - c_1 - c_2 = 1$. I haven't considered also the case $j=n-2$, but it would add a further constraint which would probably tell you exactly what is worth trying to prove.