The specific question you ask here is basically about the existence of distributional antiderivatives. The answer is "yes", and that you can also recover the "$+C$".
The main problem is this: not all test functions $\phi$ arise as a derivative of another test function. A necessary and sufficient condition is that $\int \phi = 0$. This means that the requirement
$$ \langle x,\psi'\rangle = \langle v,\psi\rangle \tag{*}\label{eq:basic}$$
for a given distribution $v$, at best, only defines what $x$ should be on a codimension-1 subspace of the space of test functions. Hence necessarily there will be some freedom.
Uniqueness
Suppose $x$ and $\hat{x}$ both solve \eqref{eq:basic}, then we have that $x - \hat{x}$ is such that, for any $\phi$ with $\int\phi = 0$ (and hence there is some $\psi$ such that $\phi = \psi'$)
$$ \langle x - \hat{x},\phi\rangle = \langle v,\psi\rangle - \langle v,\psi\rangle = 0 $$
Now as the set $\{\int \phi = 0\}$ is defined by the vanishing of a linear functional (the mapping $\phi \mapsto \int \phi$ is a distribution), this means that $x - \hat{x}$ must be a scalar multiple of the functional $\phi \mapsto \int \phi$.
In calculus notation this means that $x = \hat{x} + C$.
Existence
Fix $\eta$ a test function with $\int \eta = 1$. Given a test function $\phi$, the function $\phi - (\int\phi)\cdot \eta$ has total integral $0$, and hence we can find a test function $\psi$ such that $\psi' = \phi - (\int\phi)\cdot \eta$.
I claim that the formula
$$ \langle x, \phi \rangle = \langle v, \psi\rangle $$
defines a distribution. Linearity is obvious. It suffices to prove continuity.
Suppose $\phi$ has support in $[a,b]$, and $\eta$ has support in $[\alpha,\beta]$. Then we know $\psi$ is supported in $[\min(a,\alpha),\max(b,\beta)]$, and we have, from basic calculus, that for $s\in [\min(a,\alpha),\max(b,\beta)]$
$$ |\psi(s)| \leq (|b-a| \sup_{[a,b]} |\phi|) \cdot (1 + | \beta - \alpha| \sup |\eta|) $$
the higher derivative bounds are easier
$$ |\psi^{(k)}(s)| \leq \sup_{[a,b]} |\phi^{(k-1)}| + |b-a| (\sup_{[a,b]} |\phi|) \sup_{[\alpha,\beta]} |\eta^{(k-1)}| $$
Now let $[a,b]$ be fixed. For any $\phi$ with support in this interval, we have $\psi$ has support in $[\min(a,\alpha),\max(b,\beta)]$, and hence there exists a constant $C$ and a number $n$ (assumed to be $\geq 1$ for convenience) such that
$$ |\langle v,\psi \rangle| \leq C \sup \{ |\psi^{(k)}(s)| : s \in [\min(a,\alpha),\max(b,\beta)], k \leq n \} $$
and using the bounds above, we see that
$$ |\langle x,\phi \rangle| \leq C' \sup \{ |\phi^{(k)}(s)| : s \in [a,b], k \leq n-1 \} $$
with
$$ C' = C \cdot \big[ 1 + |b-a| \big( 1 + |\beta - \alpha| \sup |\eta| + \sup_{s\in [\beta-\alpha], k \leq n-1} |\eta^{(k)}(s)| \big) \big] $$
Note that since $\eta$ is fixed, the values $\alpha, \beta$ and the $C^k$ norms of $\eta$ are considered fixed constants.
Explicit example
In your case, fix $\eta$ a test functions with unit mass that is supported on the interval $[1,2]$, then we know that $\eta$ is the derivative of some smooth function $H$ that is identically $0$ on $(-\infty,1]$ and identically $1$ on $[2,\infty)$.
Similarly, given an arbitrary test function $\phi$ with compact support $[a,b]$, we know that it can be written as the derivative of some function $\Phi$ that is identically $0$ on $(-\infty,a]$ and $\int\phi$ on $[b,\infty)$.
So $\psi = \Phi - (\int\phi) H$ is a test function. Observe that
$$ \langle \delta_0 + 1,\psi\rangle = \Phi(0) + \int_a^{\max(b,2)} \Phi - \int \phi \cdot \int_1^{\max(b,2)} H $$
On the other hand, integrating by parts we find
$$ \int_{\leq 0} t \phi(t) + \int_{\geq 0} (1 + t)\phi(t) =
\int_a^b t \Phi' + \int_0^b \Phi' = (b+1) \int \phi - \int_a^b \Phi - \Phi(0) $$
So if we set $\hat{x}$ to be the function that equals $-t$ when $t \leq 0$ and $-t -1$ when $t > 0$, we find
$$ \langle \delta_0 + 1,\psi\rangle - \langle \hat{x},\phi\rangle =
\int_b^{\max(b,2)} \Phi + \int\phi \cdot \Big(b+1 - \int_1^2 H - \int_2^{\max(b,2)} H \Big) $$
We simplify to
$$ = \int\phi \cdot \Big( |\max(b,2)-b| + b+1 - \int_1^2 H - |\max(b,2)-2| \Big) = \int\phi \cdot \Big( 3 - \int_1^2 H \Big) $$
showing that the distributions
$$ \phi \mapsto \langle \delta_0 + 1, \psi\rangle \text{ and } \phi \mapsto \langle \hat{x},\phi\rangle $$
different by a constant ($3 - \int_1^2 H$) times the distribution $\phi \mapsto \int \phi$, exactly as expected from the discussions above.
