Definition of observer and time measured by different observers in general relativity
Your conclusion is correct, because what you are doing by saying that $g(v_{\gamma,\gamma(\lambda)},v_{\gamma,\gamma(\lambda)}) = 1$ is that the parameter $\lambda$ is exactly equal to proper time. You can have different parametrizations $\tilde{\lambda}$ of the curve $\gamma$ that have $g(v_{\gamma,\gamma(\tilde{\lambda})},v_{\gamma,\gamma(\tilde{\lambda})}) \neq 1$ and then, of course, they do not correspond to proper time of the observer on the curve.
Your conclusion from the OP just states that if you have two curves parametrized by proper time, then when they are evolved for the same amount of proper time, the same amount of proper time passes on them. A quite tautological statement!
As concerns your edit:
When you state $$p = \gamma(\lambda_1) = \delta(\lambda_1)$$ $$q = \gamma(\lambda_2) = \delta(\lambda_2)$$ and impose the normalisation of the tangents $g(v_{\gamma,\gamma(\lambda)},v_{\gamma,\gamma(\lambda)}) = g(v_{\delta,\delta(\lambda)},v_{\delta,\delta(\lambda)}) = 1$, you are NOT choosing two general curves (worldlines) passing through $p,q$. Instead, you are choosing two curves for which the same amount of proper time passes between $p,q$.
Perhaps you should consider a concrete example. Choose a space-time such as Minkowski and two randomly chosen curves $\gamma, \delta$ on it that have an arbitrary parametrization $\gamma(\tilde{\lambda}_\gamma), \delta(\tilde{\lambda}_\delta)$ and that meet at some events $p,q$. Now find the proper-time parametrisation $\lambda$ along the entire curves $g(v_{\gamma,\gamma(\lambda)},v_{\gamma,\gamma(\lambda)}) = g(v_{\delta,\delta(\lambda)},v_{\delta,\delta(\lambda)}) = 1$, this can be recast as two first-order differential equations for the functions $\lambda(\tilde{\lambda}_{\gamma,\delta})$. Each of the two differential equations have a unique solution up to a single integration constant. By setting $p = \gamma(\lambda_1) = \delta(\lambda_1)$, you specify both of these integration constants. So, you do not have any freedom to set the same condition for $\lambda_2$ at $q$ because there is no freedom in the solution left and the value of $\lambda$ at $q$ will be generally different for each of the curves. $\blacksquare$