Correction
I withdraw the previous answer (below) as wrong. @Glen_b is correct, if the second sample is not based on any results in the first, then the overall probability of selection is $\dfrac{n}{N}$, the same as for a SRS (proof below). Further, you can treat the result as a SRS although the size was not fixed in advance.
However, you stated that you took the second sample because you required more precision. If you made the decision solely because standard errors for estimates were larger than expected, then bias in the final estimated standard errors will be small (Hansen, Hurwitz, and Madow (1953, pp.77-80) (provided that the initial SE themselves were precisely estimated). In that case, I would go ahead and treat the observations as having come from a SRS. If, however, you took the second sample to reduce a borderline p-value to a smaller value, then serious bias is possible.
Proof that probability of selection is n/N:
Let $p_1 =\dfrac{n_1}{N}$ be the probability of selection in the first sample, and let $p_2$ be the conditional probability of selection in the second sample: $p_2 =\dfrac{n_2}{N-n_1}$. Let $n = n_1 + n_2$ be the final sample size. Then the probability of selection of an observation to the final sample is:
\begin{align}
p & = p_1 + (1-p_1)\thinspace p_2 \\
& = \frac{n_1}{N} + (1-\frac{n_1}{N}) \frac{n_2}{N-n_1} \\
& = \frac{n_1}{N} + \frac{N-n_1}{N}\frac{n_2}{N-n_1} \\
& = \frac{n_1 + n_2}{N} \\
& = \frac{n}{N}
\end{align}
Reference: Hansen, MH, WN Hurwitz, and W Madow. 1953. Sample Survey Methods and Theory. Volume I Methods and Applications. New York: Wiley.
Original Answer:
This kind of problem frequently arises in practice and the solution is similar.
As @David Z implies, you must investigate the possibility of systematic differences between the first and second surveys and samples.
The resulting combined sample can indeed be considered random, but not simple random. The probabilities of selection differ between the two samples, so the analysis will have to be weighted. You compute weights as follows:
Let the number in the population be $N$. Then the probability of selection for the original sample is:
$$
f_1 = \frac{107}{N}
$$
To be selected in the second sample, one must first not be selected in the first sample; then be selected in the second:
$$
f_2 = (1 - f_1) \frac{50}{N - 107}
$$
The sample weights will be $W = \dfrac{1}{f_1}$ in sample 1 and $W = \dfrac{1}{f_2}$ in sample 2.
To illustrate, suppose $N = 1020$, then the weights in the first and second sample will be $9.5327$ and $20.4$, respectively
These calculations will require modification if there was non-response in either sample.