You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process *X*_{t} − *X*_{0} = *bt* with constant *b*. In fact, you can't identify it by the n-dimensional marginals for any n.

1) Let *X* be a nondeterministic Lévy process with *X*_{0} = 0 and *n* be any positive integer. Then, there is a cadlag process *Y* with a different distribution to *X*, but such that (*Y*_{t1},*Y*_{t2},…,*Y*_{tn}) has the same distribution as (*X*_{t1},*X*_{t2},…,*X*_{tn}) for all times *t*_{1},*t*_{2},…,*t*_{n}.

Taking *n* = 2 will give a process whose increments have the same distribution as for *X*.

The idea (as in my answer to this related question) is to reduce it to the finite-time case. So, fix a set of times 0 = *t*_{0} < *t*_{1} < *t*_{2} < … < *t*_{m} for some *m* > 1.
We can look at the distribution of *X* conditioned on the ℝ^{m}-valued random variable *U* ≡ (*X*_{t1},*X*_{t2},…,*X*_{tm}). By the Markov property, it will consist of a set of independent processes on the intervals [*t*_{k−1},*t*_{k}] and [*t*_{m},∞), where the distribution of {*X*_{t} }_{t ∈[tk−1,tk]} only depends on (*X*_{tk−1},*X*_{tk}) and the distribution of {*X*_{t} }_{t ∈[tm,∞)} only depends on *X*_{tm}. By the disintegration theorem, the process *X* can be built by first constructing the random variable *U*, then constructing *X* to have the correct probabilities conditional on *U*. Doing this, the distribution of *X* at any one time only depends on the values of at most two elements of *U* (corresponding to *X*_{tk−1},*X*_{tk}). The distribution of *X* at any set of *n* times depends on the values of at most 2*n* values of *U*.

Choosing *m* > 2*n*, the idea is to replace *U* by a differently distributed ℝ^{m}-valued random variable for which any 2*n* elements still have the same distribution as for *U*. We can apply a small bump to the distribution of *U* in such a way that the *m* − 1 dimensional marginals are unchanged. To do this, we can use the following.

2) Let *U* be an ℝ^{m}-valued random variable with probability measure μ. Suppose that there exist (non-trival) measures μ_{1},μ_{2},…,μ_{m} on the reals such that μ_{1}(*A*_{1})μ_{2}(*A*_{2})…μ_{m}(*A*_{m}) ≤ μ(*A*_{1}×*A*_{2}×…×*A*_{m}) for all Borel subsets *A*_{1},*A*_{2},…,*A*_{m} ⊆ ℝ.
Then, there is an ℝ^{m}-valued random variable *V* with a different distribution to *U*, but with the same *m* − 1 dimensional marginal distributions.

By 'non-trivial' I mean that μ_{k} is a non-zero measure and does not consist of a single atom.

By changing the distribution of *U* in this way, we construct a new cadlag process with a different distribution to *X*, but with the same *n* dimensional marginals.

Proving (2) is easy enough. As μ_{k} are non-trivial, there will be measurable functions ƒ_{k} on the reals, uniformly bounded by 1 and such that μ_{k}(ƒ_{k}) = 0 and μ_{k}(|ƒ_{k}|) > 0. Replacing μ_{k} by the signed measure ƒ_{k}·μ_{k}, we can assume that μ_{k}(ℝ) = 0.
Then
$$
\mu_V = \mu + \mu_1\times\mu_2\times\cdots\times\mu_m
$$
is a probability measure different from μ. Choosing *V* with this distribution gives
$$
{\mathbb E}[f(V)]=\mu_V(f)=\mu(f)={\mathbb E}[f(U)]
$$
for any function ƒ: ℝ^{m} → ℝ^{+} independent of one of the dimensions. So, *V* has the same *m* − 1 dimensional marginals as *U*.

To apply (2) to *U* = (*X*_{t1},*X*_{t2},…,*X*_{tm}), consider the following cases.

*X* is continuous. In this case, *X* is just a Brownian motion (up to multiplication by a constant and addition of a constant drift). So, *U* is joint-normal with nondegenerate covariance matrix. Its probability density is continuous and strictly positive so, in (2), we can take μ_{k} to be a multiple of the uniform measure on [0,1].

*X* is a Poisson process. In this case, we can take μ_{k} to be a multiple of the (discrete) uniform distribution on {2*k*,2*k* + 1} and, as *X* can take any increasing nonnegative integer-valued path on the times *t*_{k}, this satisfies the hypothesis of (2).

If *X* is any non-continuous Lévy process, case 2 can be used to change the distribution of its jump times without affecting the *n* dimensional marginals: Let ν be its jump measure, and *A* be a Borel set such that ν(*A*) is finite and nonzero. Then, *X* decomposes as the sum of its jumps in *A* (which occur according to a Poisson process of rate ν(*A*)) and an independent Lévy process. In this way, we can reduce to the case where *X* is a Lévy process whose jumps occur at a finite rate, with arrival times given by a Poisson process.
In that case, let *N*_{t} be the Poisson process counting the number of jumps in intervals [0,*t*]. Also, let *Z*_{k} be the *k*'th jump of *X*. Then, *N* and the *Z*_{k} are all independent and,
$$
X_t=\sum_{k=1}^{N_t}Z_k.
$$
As above, the Poisson process *N* can be replaced by a differently distributed cadlag process which has the same *n* dimensional marginals. This will not affect the *n* dimensional marginals of *X* but, as its jump times no longer occur according to a Poisson process, *X* will no longer be a Lévy process.