Inspired by this proof in MathWorld, I rewrote the proof in terms of the Ramanujan theta function.
Define the function $$M(c)=\prod_{n = 1}^{\infty}(1 +a^{n}b^{n-1}c)\left(1+\frac{a^{n-1}b^{n}}{c}\right)\tag1$$
then $$M(abc)=\prod_{n = 1}^{\infty}(1 +a^{n+1}b^{n}c)\left(1+\frac{a^{n-2}b^{n-1}}{c}\right)\tag2$$
$$M(abc)=(1+a^2bc)\left(1+\frac{1}{ac}\right)(1+a^{3}b^{2}c)\left(1+\frac{b}{c}\right)(1+a^{4}b^{3}c)\left(1+\frac{ab^{2}}{c}\right)\cdots$$
$$M(c)=(1+ac)\left(1+\frac{b}{c}\right)(1+a^{2}bc)\left(1+\frac{ab^{2}}{c}\right)(1+a^{3}b^{2}c)\left(1+\frac{a^{2}b^{3}}{c}\right)\cdots$$
Taking $$\frac{M(abc)}{M(c)}=\left(1+\frac{1}{ac}\right)\left(\frac{1}{1+ac}\right)=\frac{1}{ac}$$
yields the following relation $$M(c)=acM(abc).$$
Now define $$N(c)=M(c)\prod_{n = 1}^{\infty}(1 -(ab)^{n}).$$
Then $$N(abc)=M(abc)\prod_{n = 1}^{\infty}(1 -(ab)^{n})$$
which becomes $$N(c)=acN(abc).$$
Now expand $N(c)$ in a Laurent series $$N(c)=\sum_{n=-\infty}^{\infty}u_{n}c^{n}.$$
Using the fundamental relation, we have $$\sum_{n=-\infty}^{\infty}u_{n}c^{n}=ac\sum_{n=-\infty}^{\infty}u_{n}(abc)^{n}$$
$$=\sum_{n=-\infty}^{\infty}u_{n}a^{n+1}b^{n}c^{n+1}$$
$$=\sum_{n=-\infty}^{\infty}u_{n-1}a^{n}b^{n-1}c^{n}.$$
Which leads to the recurrence relation $$u_{n}=u_{n-1}a^{n}b^{n-1}.$$
$$u_{1}=u_{0}a,$$ $$u_{2}=u_{1}a^{2}b=u_{0}a^{3}b,$$ $$u_{3}=u_{2}a^{3}b^{2}=u_{0}a^{6}b^{3},$$ $$u_{4}=u_{3}a^{4}b^{3}=u_{0}a^{10}b^{6}.$$
Which in general form is $$u_{n}=u_{0}a^{n(n+1)/2}b^{n(n-1)/2}.$$
Now substituting back into the original Laurent series, we obtain $$N(c)=u_{0}\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$
It can easily be shown that $u_{0}=1$, so that we have the Jacobi triple product in terms of the Ramanujan theta function $$N(c)=\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$
Q:Is this generalisation of the proof correct?
Revised: 25 April 2020
Following Somos' critique, I searched around a couple of papers and finally came across the following on wikipedia
Let us set $a=b=q=e^{2\pi i \tau}$ and $c=1$ and show that the following numerator and denominator
$\frac{1}{u_{0}(e^{2\pi i n^2 \tau})}=\frac{\sum_{n=-\infty}^{\infty}e^{2\pi i n^2 \tau}}{\prod_{k = 1}^{\infty}(1-e^{4\pi i k \tau})\left(1+e^{2\pi i (2k-1) \tau}\right)^2}$
are weight 1/2 modular functions under the transformation $\tau = \frac{-1}{4 \tau}$
Since a modular function is 1-periodic $f(q+1)=f(q)$ and bounded on the upper half plane $\tau \to i \infty$, it ought to be constant according to Louiville's theorem, therefore the quotient has to be constant such that $u_{0}(q)=u_{0}(0)=1$
The proof you have is essentially the same as the MathWorld proof with a few reparametrizing of terms. It stands or falls with the one you are modeling it on. It is similar in the way that Ramanujan's theta function differs from Jacobi's theta function and I commend you for that.
Unfortunately, that proof is flawed. The crux is the step from $M(c)$ to $N(c)$ in your proof. It comes from nowhere. Why did you multiply by that infinite product? The same problem arises in the Mathworld proof going from $F(z)$ to $G(z)$. Without being able to justify $a_0=1$ the proofs are not complete. Of course, the result is still true, but the justification is missing.