Let $\mathcal{L}(X)$ indicates the law of a random variable from a probability space $\Omega$ into a Polish space $E$ (so $E=\mathbb{R}$ for a real random variable $X$, and $E=\mathbb{D}(\mathbb{R})$ is the space of all cadlag functions form $\mathbb{R}_{+}$ to $\mathbb{R}$ for a stochastic process $X$).
From Lemma VI.3.19 of Jacod and Shiryaev's book, we have the following property.
Let $X$ and $Y$ be two cadlag stochastic processes. If $$ \mathcal{L}(Y_{t_1},\dots,Y_{t_k}) = \mathcal{L}(X_{t_1},\dots,X_{t_k}) $$ for all $t_j\in D$, $D$ dense subset of $\mathbb{R}_{+}$, and for all $k\in\mathbb{N}^{\star}$, then $\mathcal{L}(X)=\mathcal{L}(Y)$.
Suppose now that $X_t\stackrel{d}{=}Y_t$ for all $t\in \mathbb{R}_{+}$. Can we still conclude that $\mathcal{L}(X)=\mathcal{L}(Y)$ ?
Addendum. Mostly, I need some help dealing with the law of a stochastic process. Formally, we have to take a set $A$ of the Borel $\sigma$-algebra on $\mathbb{D}(\mathbb{R})$ and prove that $\mathsf{P}(X^{-1}(A))=\mathsf{P}(Y^{-1}(A))$, where $X^{-1}(A)=\{\omega\in\Omega | \{X_{t}(\omega)\}_{t\geq 0}\in A\}$, where $\{X_t(\omega)\}_{t\geq 0}$ indicates the whole trajectory of $X$ (and similarly for $Y$).
No, it's not enough to verify $\mathcal L (X_t)=\mathcal L(Y_t)$ for every $t$ to conclude that $\mathcal L(X)=\mathcal L(Y)$.
For a counterexample, let $Z$ have standard normal distribution and define stochastic process $(X_t, t>0)$ via $$X_t:=\sqrt{t} Z.$$ Then each $X_t$ has normal distribution with mean $E(X_t)=0$ and variance $\operatorname{Var}(X_t)=t$. In other words, for each $t$ the variable $X_t$ has the same distribution as $W_t$ for $W$ a Brownian motion. But $X$ is decidedly not Brownian motion.