The system is equivalently LTI for shifts that are even in number of samples, or for inputs unaliased by the subsampling (zero spectrum over $[f_s/4, -f_s/4)$). It is always linear. These can be verified by working through Subsampling in time <=> Folding in Fourier, also through FFT upsampling. The $+(x * h)[n]$ part is always LTI, and sum of LTI is LTI.
The test in Python would look like:
out0 = op(np.roll(a + b, shift), d)
out1 = np.roll(op(a, d) + op(b, d), shift)
assert np.allclose(out0, out1)
where op(x, d) implements the system (no need for upper branch) with d as the subsampling factor, and roll is a circular shift. The test says, "delaying input delays output, and summing inputs sums outputs". Circularity in roll is made equivalent with linearity with sufficient zero-padding.
Purpose for signal processing, unclear. If $h$ is a high-pass with no frequencies outside $[f_s/4, -f_s/4)$, then it plants a copy of the highpassed signal onto low frequencies, while retaining the original low frequencies except lowpassed by the mirrored $\hat h$. If $\hat h$ has the opposite support (lowpass), then it doubles $(x * h)[n]$ if $x$ is unaliased, else it copies highs the same way but now the upper branch is empty in highs. For general $\hat h$, both lows and highs in the original $\hat x$ are equally affected for the subsampled branch, but only highs are duplicated in the end result. So it's a really shitty way of emphasizing high frequencies.
Purpose for deep neural networks, it resembles a skip connection in ResNets, where the idea is to constrain the flow of information to promote sparsity. However, there the two $h$ would differ. If they're same, it's useless as a nonlinearity, so the only non-triviality is in shifts - it'd be aiming for a very specific feature that I can't think of.
Also, the diagram is underspecified; it makes the most sense to also subsample $h[n]$ by 2. For LTI-ness, the conclusions either way don't change as they hold for any general $h$, and this lower compute branch is independent of the upper. The spectral analysis I've done earlier does change significantly, but I won't account for this other (I think esoteric) case, except by saying that now it's never doubling $(x * h)[n]$, nor do I see it adding purpose. Not subsampling $h$ by 2 introduces the independent problem of handling boundaries, as now $h$'s support is relatively doubled. It remains LTI for any integer "shift / sub_factor", and linear for all parameters. I also assumed the upsampling is sinc interpolation / DFT upsampling, else this answer could change entirely.
downsampling operator is not time-invariant
Subsampling isn't time-invariant, but upsampling the subsampling is, under the stated constraints. For the bandlimited case, it's faithfully LTI - for all others, it's "equivalently" so (the math coincides, though in interesting ways). This is explored more fully - with proofs and code validation - at Is downsampling LTI for bandlimited inputs?.
