Conditional expectation $\mathbb E\left(\exp\left(\int_0^tX_sdB_s\right) \mid \mathcal F_t^X\right)$

First of all, note that we have to ensure that

$$\exp \left( \int_0^t X_s \, dB_s \right) \in L^1. \tag{1}$$

If your claim is true, then

$$\mathbb{E} \exp \left( \int_0^t X_s \, dB_s \right) = \mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right),$$

i.e. $(1)$ holds if

$$\mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right)< \infty. \tag{2}$$

Through the remaining part of my answer, I'll assume that $(2)$ holds. The last part of the following proof is very close to the proof of Novikov's condition in René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.

We denote by $(M_t)_{t \geq 0}$ the stochastic exponential of $(X_t)_t$:

$$M_t := \mathcal{E}(X)_t := \exp \left( \int_0^t X_s \, dB_s- \frac{1}{2} \int_0^t X_s^2 \, ds \right).$$

Then the claim is equivalent to $\mathbb{E}(M_t \mid \mathcal{F}_t^X)=1$. Moreover, we set $$Y_t := \int_0^t X_s \, dB_s \qquad \text{and} \qquad \langle Y \rangle_t := \int_0^t X_s^2 \, ds.$$

Using e.g. the approximation procedure described in the OP and (conditional) Fatou's lemma, it is not difficult to see that $$\mathbb{E}(M_t \mid \mathcal{F}_t^X) \leq 1.$$ Consequently, it remains to show that $$\mathbb{E}(M_t \mid \mathcal{F}_t^X) \geq 1. \tag{3}$$

For $n \in \mathbb{N}$ we define

$$\tau_n := \inf\{t \geq 0; \max\{|B_t|,|X_t|\} \geq n\}.$$

Since, by Itô's formula,

$$M_{t \wedge \tau_n} -1 = \int_0^{t \wedge \tau_n} M_s \, dB_s$$

we find for any $F \in \mathcal{F}_t^X$

$$M_{t \wedge \tau_n}1_F-1_F = \int_0^{t \wedge \tau_n} (1_F M_s) \, dB_s.$$

(Note: Because of the independence of $X$ and $B$, we can put the indicator function $1_F$ under the integral sign.) As the right-hand side is a martingale, we get

$$\mathbb{E}(M_{t \wedge \tau_n} 1_F) = \mathbb{P}(F).$$

Since this holds for any $F \in \mathcal{F}_t^X$, this shows

$$\mathbb{E}(M_{t \wedge \tau_n} \mid \mathcal{F}_t^X) = 1. \tag{4}$$

In particular,

$$\mathbb{E}(M_{t \wedge \tau_n}) = 1. \tag{5}$$

For fixed $c \in (0,1)$, we pick $p=p(c)>1$ such that $p < \frac{1}{c(2-c)}$. By Hölder's inequality (for the exponents $1/pc$ and $1/(1-pc)$), we obtain

$$\begin{align*} \mathbb{E}[\mathcal{E}(c X_{t \wedge \tau_n})^p] &\stackrel{\text{def}}{=} \mathbb{E} \exp \left( pc Y_{t \wedge \tau_n} - \frac{1}{2} pc^2 \langle Y \rangle_{t \wedge \tau_n} \right) \\ &= \mathbb{E} \left[\exp \left(pc (Y_{t \wedge \tau_n} - \frac{1}{2} \langle Y \rangle_{t \wedge \tau_n} \right) \exp \left( \frac{1}{2} pc(1-c) \langle Y \rangle_{t \wedge \tau_n} \right) \right] \\ &\leq \underbrace{ \left[\mathbb{E}\exp \left( Y_{t \wedge \tau_n} - \frac{1}{2} \langle Y \rangle_{t \wedge \tau_n} \right) \right]^{pc}}_{[\mathbb{E}(M_{t \wedge \tau_n}]^{pc} \stackrel{(5)}{=} 1} \left[ \mathbb{E} \exp \left( \frac{1}{2} \frac{pc(1-c)}{1-pc} \langle Y \rangle_t \right) \right]^{1-pc} \\ &\leq \left[ \mathbb{E} \exp \left( \frac{1}{2} \frac{pc(1-c)}{1-pc} \langle Y \rangle_T \right) \right]^{1-pc} \end{align*}$$

for any $t \leq T$. This shows that the $p$-th moments of the family $(\mathcal{E}(c X_{t \wedge \tau_n}))_{n \in \mathbb{N}}$ is uniformly integrable. It follows from Vitali's convergence theorem that

$$\mathcal{E}(c X_{t \wedge \tau_n}) \to \mathcal{E}(cX_t) \qquad \text{in $L^1$} \tag{6}$$

as $n \to \infty$. By $(4)$ (applied for $\tilde{X} := cX$), this implies

$$\begin{align*} 1 &\stackrel{(4)}{=} \lim_{n \to \infty} \mathbb{E}\left[ \exp \left( cY_{t \wedge \tau_n} - \frac{1}{2} c^2 \langle Y \rangle_{t \wedge \tau_n} \right) \mid \mathcal{F}_t^X \right] \\ &\stackrel{(6)}{=} \mathbb{E}\left[ \exp \left( cY_{t} - \frac{1}{2} c^2 \langle Y \rangle_{t} \right) \mid \mathcal{F}_t^X \right] \end{align*}$$

Finally, we use a last time (conditional) Hölder's inequality (for $1/c$ and $1/(1-c)$) to obtain

$$\begin{align*} 1 &= \mathbb{E} \left[ \exp \left( c Y_t - \frac{1}{2} c \langle Y \rangle_t \right) \exp \left( \frac{1}{2} c(1-c) \langle Y \rangle_t \right) \mid \mathcal{F}_t^X \right] \\ &\leq \left[ \mathbb{E} \left( \exp \left( Y_t- \frac{1}{2} \langle Y \rangle_t \right) \mid \mathcal{F}_t^X \right) \right]^c \left[ \mathbb{E} \left( \exp \left( \frac{1}{2} \langle Y \rangle_t \right) \mid \mathcal{F}_t^X \right) \right]^{1-c} \end{align*}$$

Since, by $(2)$, the last term is bounded for $t \leq T$, we can let $c \uparrow 1$:

$$1 \leq \mathbb{E} \left[ \exp \left( Y_t - \frac{1}{2} \langle Y \rangle_t \right) \mid \mathcal{F}_t^X \right] \stackrel{\text{def}}{=} \mathbb{E}(M_t \mid \mathcal{F}_t^X).$$