Just for fun: assuming that you (or your compiler) vectorize both of your approaches (because of course you would if you're chasing performance), and you're targeting a recent x86 processor...
A direct translation of "LerpColorSolution1" into AVX instructions is as follows:
VSUBPS dst, a, b // a[] - b[]
VSHUFPS alpha, alpha, alpha, 0 // splat alpha
VMULPS dst, alpha, dst // alpha*(a[] - b[])
VADDPS dst, a, dst // a[] + alpha*(a[] - b[])
The long latency chain for this sequence is sub-mul-add, which has a total latency of 3+5+3 = 11 cycles on most recent Intel processors. Throughput (assuming that you do nothing but these operations) is limited by port 1 utilization, with a theoretical peak of one LERP every two cycles. (I'm intentionally overlooking load/store traffic and focusing solely on the mathematical operation being performed here).
If we look at your "LerpColorSolution2":
VSHUFPS alpha, alpha, alpha, 0 // splat alpha
VSUBPS dst, one, alpha // 1.0f - alpha, assumes "1.0f" kept in reg.
VMULPS tmp, alpha, b // alpha*b[]
VMULPS dst, dst, a // (1-alpha)*a[]
VADDPS dst, dst, tmp // (1-alpha)*a[] + alpha*b[]
Now the long latency chain is shuffle-sub-mul-add, which has a total latency of 1+3+5+3 = 12 cycles; Throughput is now limited by ports 0 and 1, but still has a peak of one LERP every two cycles. You need to retire one additional µop for each LERP operation, which may make throughput slightly slower depending on the surrounding context.
So your first solution is slightly better; (which isn't surprising -- even without this detail of analysis, the rough guideline "fewer operations is better" is a good rule of thumb).
Haswell tilts things significantly in favor of the first solution; using FMA it requires only one µop on each of ports 0,1, and 5, allowing for a theoretical throughput of one LERP per cycle; while FMA also improves solution 2, it still requires four µops, including three that need to execute on port 0 or 1. This limits solution 2 to a theoretical peak of one LERP every 1.5 cycles -- 50% slower than solution 1.