Generalizing Ramanujan's proof of Bertrand's Postulate: Can Ramanujan's approach be used to show a prime between $4x$ and $5x$ for $x \ge 3$
Ramanujan's proof is actually a simplification of Chebyshev's original [1852] proof of Bertrands postulate (the article is Memoire sur les nombres premiers. J. Math. Pures Appl. 17 1852). Chebyshev uses a stronger approach proving a lot more than Bertrand's postulate, in particular your statement follows directly from his bound while his methods are essentially the same as Ramanujan's
He starts deriving the identity $$ \log [x]! = \psi(x) + \psi\left(\frac x2 \right) + \psi\left(\frac x3 \right) + \dots $$ and then he uses it to derive the following identity (similar to Ramanujan's or your's but stronger):
$$ \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } = \psi\left(x\right)-\psi\left(\frac{x}{6}\right)+\psi\left(\frac{x}{7}\right) -\psi\left(\frac{x}{10}\right)+\psi\left(\frac{x}{11}\right) -\psi\left(\frac{x}{12}\right)+\psi\left(\frac{x}{13}\right) -\psi\left(\frac{x}{15}\right)+\psi\left(\frac{x}{17}\right) -\psi\left(\frac{x}{18}\right)+\psi\left(\frac{x}{19}\right) -\psi\left(\frac{x}{20}\right)+\psi\left(\frac{x}{23}\right) -\psi\left(\frac{x}{24}\right)+\psi\left(\frac{x}{29}\right) -\psi\left(\frac{x}{30}\right) + \dots - \dots $$ where the sequence in the right continues with period 30 in he denominator, ie the first missing terms are $\psi(x/31)-\psi(x/36)+\psi(x/37)- \dots $.
As you can see this is similar to the argument you give above, as we have an alternating sequence of non-increasing terms, so you can bound it above and below stopping after an odd/even number of terms: $$ \psi(x) - \psi\left(\frac x6\right) \le \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } \le \psi(x) $$
Now he uses Stirling to find the approximation $$ Ax - \frac{5}{2}\log x - 1 < \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } < A x + \frac{5}{2}\log x $$ where $$ A = \log \frac{2^{1/2}3^{1/3}5^{1/5}}{30^{1/30}} = 0.92129202 $$ which combined with the previous inequalities gives: $$ \psi(x) > Ax -\frac{5}{2}\log x -1 \quad\text{and}\quad \psi(x)-\psi\left(\frac{x}{6}\right) < Ax + \frac{5}{2}\log x $$ He uses an auxiliary funciont to telescope this inequality and finds that: $$ \psi(x) < \frac{6}{5}Ax + \frac{5}{4\log 6} \log^2 x + \frac{5}{4}\log x + 1 $$ And now he uses the inequality $$ \psi(x) - 2\psi(\sqrt{x}) < \theta(x) < \psi(x) - \psi(\sqrt{x}) $$ to derive (after some technical work) that there are more than $k$ primes between $l$ and $L$ if $$ l = \frac{5}{6} L - 2 \sqrt{L} - \frac{25 \log^2 L}{16\log 6 A}-\frac{5}{6A}\left(\frac{25}{4}+k\right)\log L - \frac{25}{6A} $$ So with $k = 0$ we find that there is a prime between $4x$ and $5x$ for $x >= 2034$ and we can check numerically that the same holds for $x \ge 1331$.
(However this falls short to prove that there is always a prime between $5x$ and $6x$.)
Some comments about your clarifications:
As it has been commented there is an elementary proof of the prime number theorem so there is an elementary proof that there is a prime between $x$ and $(1 + \epsilon) x$ for all $\epsilon$.
This is the first result of his kind, but the bounds obtained by Chebyshev are not easy to improve without using the PNT. First Sylvester [1881] On Tchebycheff's theory of the totality of the prime numbers comprised within given limits. Improved the upper bound on $\psi(x)$ to $$ 0.95695 x \le \psi(x) \le 1.04423 x $$ Using (I think) combinations of identities as Chebyshev's and Ramanujan's. I never was able to find the article so I'm not sure about Sylvester's method or results (so please check them). With these inequalities it should be possible to prove the existence of a prime between $10x$ and $11x$ for large enough $x$.
In http://arxiv.org/pdf/0709.1977v1.pdf you can find a list of all possible identities that can be used almost directly in Chebyshev's method, there are several infinite families and 52 isolated identities. (however I believe Chebyshev's identity is the one that gives the best possible bound using only his method).
I think using Rosser and Schoenfeld's result $\psi(x)<1.03883x$ is (in some sense) a major gap in your argument. I don't know how it is derived but I think as there are a lot of results in that article that imply the existence of a prime between $kx$ and $(k+1)x$ for every prime, so I would avoid it to prevent circular reasoning.