In fact, like other SR experts, you forgot how the LT
have been derived.
In particular, time is counted from the instant at
which the origins of the two frames of reference S and S'
coincide. Iow, t=t'=0 when x=x'=0.
If a light pulse or an object starts from the coincident
origins at time t=t'=0, it will be at x=ct or x=Vt according
to S, and at x'=ct' or x'=V't' according to S'.

The classical LT x'=gamma(x-vt) and t'=gamma(t-vx/c^2) can be
written x'=gamma*t(V-v) and t'=gamma*t(1-Vv/c^2), where x=Vt.
Clearly, t' can be negative only if V>c, which is forbidden
by relativity.

Or the transformation t'=gamma(t-vx/c^2), applied to the event
x=a, t=0, gives t'= -gamma va/c^2, thus a negative time.
Consequently, the transformation is not generally applicable,
the Einsteinian space-time is physically wrong, and SRT itself
is definitely wrong.

No wonder that the classical LT, based on x=Vt, give false
results when x=a+Vt, which is the case of the event x=a, t=0.

In order to be applied to all possible cases, the LT must be
generalized to x'=gamma(x-vt) and t'=gamma(t-v(x-a)/c^2).
With x=a and t=0, x-a is of course 0, x'=gamma*a and
t'=0, a correct result.

The generalized LT can also be written
x'=gamma*a + gamma*t(V-v) and t'=gamma*t(1-Vv/c^2).

Transformed to the frame S', this equation becomes:
 x' = xo + ut'
where xo = sqrt(1-v^2/c^2)/(1 - vV/c^2)
and   u = (V-v)/(1 - vV/c^2)

So the position of the object in S' at t' = 0 is xo."

Your x0 is false. As a=1, it should be 1/sqrt(1-v^2/c^2).
Note also that the correct LT doesn't lead to silly infinities
any more.